text stringlengths 1 2.12k | source dict |
|---|---|
If c = 1 this means T(n) < 1 7/8 n. Not what we need, we need T(n) < cn. If c = 2 it means T(n) + 2 3/4 n. Slightly better but not good enough. For which c is (7/8 c + 1) n ≤ cn, or 7/8 c + 1 ≤ c? That's the case for c ≥ 8. The start of the induction is n=0, and it's easy to show that T(0) = 0.
So you haven't just shown that T(n) = O(n), you have shown the much stronger T(n) ≤ 8n.
• Thank you this is an excellent answer and helps me understand these kinds of problems in more depth! – Mandy Mar 5 at 23:46
• You might have forgotten the base case. – Yuval Filmus Mar 6 at 16:38 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9835969698879862,
"lm_q1q2_score": 0.8597397102769623,
"lm_q2_score": 0.8740772253241803,
"openwebmath_perplexity": 335.2118537186417,
"openwebmath_score": 0.7982414960861206,
"tags": null,
"url": "https://cs.stackexchange.com/questions/121468/proving-that-tn-t-lfloor-n-2-rfloor-t-lfloor-n-4-rfloor-t-lfloor"
} |
### NBHM 2020 PART A Question 10 Solution: Possible rank of a matrix obtained by changing the entries $(1, 3)$, $(1, 4)$, $(3, 3)$, $(3, 4)$, and $(5, 4)$
Let $M$ be a $7×6$ real matrix. The entries of $M$ in the positions $(1, 3)$, $(1, 4)$, $(3, 3)$, $(3, 4)$, and $(5, 4)$ are changed to obtain another $7×6$ real matrix $\widetilde{M}$. Suppose that the rank of $\widetilde{M}$ is 4. What could be the rank of $M$? List all possibilities.
Solution:
First, we will derive a connection between the rank of $M$ and the rank of $\widetilde{M}$. We have,
$$rank(M) = rank(M-\widetilde{M}+\widetilde{M}) \le rank(M-\widetilde{M})+rank(M)$$ Similarly
$$rank(\widetilde M) = rank(\widetilde{M}-M+M) \le rank(\widetilde{M}-M)+rank(M)$$
This shows that $rank(\widetilde{M}) - rank(\widetilde{M}-M) \le rank(M) \le rank(M-\widetilde{M})+rank(\widetilde{M})$
The matrices $M - \widetilde{M}$ and $\widetilde{M}-M$ can have non-zero entries only in the columns $3$ and $4$ and hence they are of rank atmost $2$. Also, it is given that the rank of $\widetilde{M}$ is equal to $4$. Hence we have $$2 \le rank(M) \le 6.$$
Using $0,1$ matrics one can check that all these possibilities are occurring indeed.
Share to your groups:
FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS AND COMMENTS BELOW IN THE COMMENTS SECTION. ALSO, SUGGEST PROBLEMS TO SOLVE.
### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft... | {
"domain": "theindianmathematician.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9915543740571681,
"lm_q1q2_score": 0.8597130915789817,
"lm_q2_score": 0.8670357512127872,
"openwebmath_perplexity": 162.1564447633871,
"openwebmath_score": 0.7001168131828308,
"tags": null,
"url": "https://nbhmcsirgate.theindianmathematician.com/2020/03/nbhm-2020-part-question-10-solution.html"
} |
# More on 0.999…
#### (An archive question of the week)
In collecting questions and answers about 0.999… for the last post, there were two that were too long to include, but that dig more deeply into issues that some of the standard answers tend to gloss over. So here, I want to look at those two answers, both of which deal with how to handle arithmetic on infinitely many digits: How can you add, subtract, or multiply when there is no rightmost digit? This is not central to the main question, but enough people express uncertainty about this aspect to make it worth covering.
## Adding without a place to start
Here’s the first question, from Darryl in 1999 (appropriately enough):
What is 0.999... + 0.999...?
While discussing the 1 = 0.999... solution, a person asked what is 0.999... + 0.999...? I think it is a good question - one that I could not answer. It should be 2 but how do we show this?
In the archives at
Getting 0.99999...
http://mathforum.org/dr.math/problems/dusty4.15.98.html
you say "Can you figure out why 0.3bar + 0.3bar = 0.6bar? Because these numbers go on forever, you will need to use a little logic to add them. (The algorithm that you learned for adding numbers doesn't work very well when you can't get to the rightmost number.)"
If we copy the idea of limits, we could say that for any number n we could start adding at this rightmost point and get the 0.66...6(nth) place. And since we can do this for any n then the equation holds. But if we try this for 0.9bar + 0.9bar we seem to get something that looks less like 2, namely 1.9....8(nth). But if we replace 8 with 9, we seem to get a number between 2 and the one we have, which makes me feel as if 0.9bar + 0.9bar < 2. | {
"domain": "themathdoctors.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9854964194566753,
"lm_q1q2_score": 0.8596952109972292,
"lm_q2_score": 0.8723473713594992,
"openwebmath_perplexity": 641.1596823095123,
"openwebmath_score": 0.7628352046012878,
"tags": null,
"url": "https://www.themathdoctors.org/more-on-0-999/"
} |
The link is to an elaborated version of the 1/3 + 1/3 + 1/3 = 1 argument, which I only gave a link to last time. (The notation 0.3bar as a typable version of $$0.\overline{3}$$, comes from there.) As Doctor Derrel pointed out there, we normally add from right to left, and there is no rightmost digit to start at in 0.333… + 0.333…; fortunately, we know that it isn’t necessary to start at the right when there are no carries, so we can just add left to right. This is discussed here:
Darryl’s suggestion is to consider any n digits, where we can clearly add 0.333…3 + 0.333…3 = 0.666…6, so we get the desired result by letting n increase without bound.
But, as Darryl points out, things get trickier when you try to add 0.999… + 0.999… . With 10 decimal places, for example, we get
0.9999999999
+ 0.9999999999
--------------
1.9999999998
You get a carry into each place from the place to its right, but where can you start that? And if there’s an 8 anywhere, then we’re less than 1.999… = 2, right? The answer, ultimately, is that in fact the 8 is nowhere!
Hi Darryl, thanks for your question. It's an interesting twist on the perennial debate about whether 0.999... could really equal 1.
Since that 8 is the nth digit and you take n to infinity, there is no digit 8 in the limit.
With no rightmost place where you would have 9 + 9 = 18, every digit will be 9. But that’s just impossible to imagine. So we have to do as mathematicians do, and find a way to work around infinities. That way is to use limits formally.
Apply the definition of a limit to what you have. We can express it as a game: you pick a number epsilon, as small as you want, and I have to pick a number n such that
|2 - 1.999...98| < epsilon
where the 8 is in the nth decimal place. Since
|2 - 1.999...98| = 2*10^(-n)
I just pick
n > -log(epsilon/2) | {
"domain": "themathdoctors.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9854964194566753,
"lm_q1q2_score": 0.8596952109972292,
"lm_q2_score": 0.8723473713594992,
"openwebmath_perplexity": 641.1596823095123,
"openwebmath_score": 0.7628352046012878,
"tags": null,
"url": "https://www.themathdoctors.org/more-on-0-999/"
} |
|2 - 1.999...98| = 2*10^(-n)
I just pick
n > -log(epsilon/2)
This is no harder than in the case of 0.999...9. The 8 in the last place has no effect on the limit; it only delays somewhat the approach to the limit, as indicated by that factor of 1/2.
The idea here is that we can consider such a sum with any number n of decimal places, which will have an 8 in the last place. We can get this sum to be as close as we want to 2, just by taking enough decimal places; in effect, to get as close as we want, we just have to take enough digits to push the 8 beyond where it would make the error too large. If we want our sum to be less than $$\epsilon = 0.00000000000000000001$$ away from 2, we just need to use $$n > -\log(\epsilon/2) = -\log(0.00000000000000000001/2) = 20.3$$, so 21 digits will be enough: that is, 0.999999999999999999999 + 0.999999999999999999999 = 1.999999999999999999998.
Here is another way to approach the problem, by manipulating infinite sums. If you want to make everything explicit, you can write the sums as limits of finite sums, and get into the issue of switching the limit and the sum as you go through this process.
Let's write 0.999... as an infinite sum:
0.999... = Sum[n = 1 to infinity](9*10^-n)
Now we can add infinite sums:
0.999... + 0.999...
= Sum[n = 1 to infinity](9*10^-n)
+ Sum[n = 1 to infinity](9*10^-n)
= Sum[n = 1 to infinity](18*10^-n)
= Sum[n = 1 to infinity](10*10^-n + 8*10^-n)
= Sum[n = 1 to infinity](10^(-n+1))
+ Sum[n = 1 to infinity](8*10^-n)
= Sum[n = 0 to infinity](10^-n)
+ Sum[n = 1 to infinity](8*10^-n)
= 1 + Sum[n = 1 to infinity](10^-n)
+ Sum[n = 1 to infinity](8*10^-n)
= 1 + Sum[n = 1 to infinity]((8+1)*10^-n)
= 1 + Sum[n = 1 to infinity](9*10^-n)
= 1.999...
I think this is the result you are looking for. Since we know that 0.999... = 1, the answer is 1 + 1 = 2; but you wanted to see that unbroken string of 9's. | {
"domain": "themathdoctors.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9854964194566753,
"lm_q1q2_score": 0.8596952109972292,
"lm_q2_score": 0.8723473713594992,
"openwebmath_perplexity": 641.1596823095123,
"openwebmath_score": 0.7628352046012878,
"tags": null,
"url": "https://www.themathdoctors.org/more-on-0-999/"
} |
He is adding an infinite string of carries (1) to an infinite string of 8’s, and getting an infinite string of 9’s.
By the way, notice that the final 8 never appears in this method. That's because we're dealing with infinite sums from the start, so there is always a carry from the next digit to the right. The sum of 10^-n is the sum of the infinite number of carries.
## Multiplying without a place to start
A similar question arises in thinking about the standard method of conversion from a repeating decimal. This question is from 2000:
Induction on .999...
I have been having an argument with my physics teacher over the fact that point nine recurring (.999... or PNR) equals 1. I showed him the proof on your site and he pointed out a fact that I hadn't noticed.
How can you multiply PNR by ten if you can't get to the beginning? For example, to multiply 215 * 3, you would start off on the right, multiplying 3 by 5, then 1, then finally 2. Now how can you multiply PNR by ten if the farthest right value cannot be reached?
Thank you for any clarification you can give.
As you should recall, we convert 0.999… to a fraction by subtracting it (x) from 10x:
x = 0.9999...
10x = 9.9999...
10x - x = 9.0000...
9x = 9
x = 1
Here, we first had to multiply 0.9999… by 10. But, as in the addition case above, there is no rightmost digit at which to start the multiplication, so can we actually do that multiplication? That is Blake’s teacher’s challenge.
Doctor TWE took it up, first referring to the FAQ (which includes links to several of the answers I’ve referred to) as a source of supplemental arguments before turning to the main issue. | {
"domain": "themathdoctors.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9854964194566753,
"lm_q1q2_score": 0.8596952109972292,
"lm_q2_score": 0.8723473713594992,
"openwebmath_perplexity": 641.1596823095123,
"openwebmath_score": 0.7628352046012878,
"tags": null,
"url": "https://www.themathdoctors.org/more-on-0-999/"
} |
I wanted to comment on your (or your teacher's) idea that "you can't get to the beginning." We multiply values from right to left only as a matter of convenience, so we don't have to "backtrack" each time we carry. It is equally valid to multiply from left to right. This is often done when we want to get a quick estimate for the answer, then get more accuracy later. The LEFTMOST digits contribute the most to the answer, so to get "in the ballpark," we can multiply them first. I'll demonstrate with a concrete example, then a general argument.
We’ve discussed this left-to-right idea in Dividing Right to Left, Adding Left to Right.
Let's use your example of 215 * 3. I can multiply them as follows; first, I'll multiply the 3 by the 2 (actually, by 200) and put the answer in the hundreds place, like this:
215
* 3
---
6
Then I'll multiply the 3 by 1 and put the answer in the tens place:
215
* 3
---
6
3
Finally, I'll multiply the 3 by 5 and put the answer in the units place:
215
* 3
---
6
3
15
215
* 3
---
6
3
15
---
645
Notice that along the way, I got pretty good estimates for my final answer. After the first step my total was 600 (not a bad estimate, off by less than 10%), after the second step I had 630, and after the third step, I got 645. Going right to left, my totals after each step would be: 15 (not a very good estimate of the final answer), 45 (still not very good) and finally 645.
So multiplying left to right just requires us to modify the digits we already have to account for carries from new columns as we do them; but the effect of the new column rarely propagates very far back into our work, and never makes a large change. Each new digit makes a smaller adjustment than the one before. That will be relevant as we continue.
In the general case, consider multiplying a 3-digit value ABC by some value X, where A, B and C are digits. What we really have, then, is
(100*A + 10*B + 1*C) * X | {
"domain": "themathdoctors.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9854964194566753,
"lm_q1q2_score": 0.8596952109972292,
"lm_q2_score": 0.8723473713594992,
"openwebmath_perplexity": 641.1596823095123,
"openwebmath_score": 0.7628352046012878,
"tags": null,
"url": "https://www.themathdoctors.org/more-on-0-999/"
} |
(100*A + 10*B + 1*C) * X
Conventionally, we solve this starting with the units digit as:
X*C*1 + X*B*10 + X*A*100
But using the commutative property of addition, this is equal to:
X*A*100 + X*B*10 + X*C*1
showing that the order doesn't matter. If I wanted to, I could start in the middle, for example:
X*B*10 + X*A*100 + X*C*1
I'd just have to be careful not to miss any digits or do any ones twice.
What we see here is that, conceptually, we can think of the whole multiplication as being done at once; it doesn’t matter in what order we do the pieces, because the sum is commutative.
Without being able to start with the most significant digit, we could never find a value like 2*pi, because pi (like all irrationals) is an infinite non-repeating decimal. When we say 2*pi is approximately 6.2831853, we can do so because we started multiplying at the end and not the beginning.
One final note: How can we be sure that the multiplications by 10 continue as we expect (i.e. they continue shifting the digits 1 place to the left) and that the subtractions of successive digits produce zeroes infinitely? These steps can be proven using a technique called mathematical induction. That's too complex to explain here, but if you search our Ask Dr. Math archives for the word "induction" (type it without the quotes), you'll find many questions and answers about it.
Whenever we do a calculation on numbers like π in a calculator, we are working only with the leftmost digits (those the calculator can hold); but since we know that the digits we have omitted will add no more than a 1 in the rightmost digit by carrying, the answer will be as close as we need. (In fact, calculators actually work with another digit or two beyond what they display, so that such errors never even show up.)
## Proof by induction | {
"domain": "themathdoctors.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9854964194566753,
"lm_q1q2_score": 0.8596952109972292,
"lm_q2_score": 0.8723473713594992,
"openwebmath_perplexity": 641.1596823095123,
"openwebmath_score": 0.7628352046012878,
"tags": null,
"url": "https://www.themathdoctors.org/more-on-0-999/"
} |
## Proof by induction
That last teaser – that there is a formal method for actually proving infinite things without having to do infinite things – was just what Blake needed. He replied a couple days later (back then we didn’t have threaded conversations, so he didn’t know whether the same Doctor would get the message):
I recently sent you a letter regarding the proof that .999... = 1 in the FAQ. Specifically, I asked how it was possible to multiply PNR by ten when you could not get to the right-hand side of it. This was explained quite neatly.
However, the nice person who helped me out said it could be proven that one can multiply PNR by 10 by using a process of induction. I tried to search the archives, but the results I found there were not very satisfactory because it seemed to me that each induction question was actually problem-specific.
To cut short, I would like to know the induction proof for multiplying PNR by ten.
Doctor TWE first summarized the concept of induction:
Hi again Blake! Thanks for writing back!
In general, we use proof by induction whenever we want a proof that involves an infinite sequence or series, in this case .999...
Inductive proofs require two steps:
Step 1 (the basis step): Prove it for some starting value, like n = 1.
Step 2 (the inductive step): Prove that if it's true for n = k, then it is true for n = k+1.
For the inductive step, we assume that it is true for n = k, and we usually use this assumption in the proof itself.
A word of caution: because of their self-referential nature, inductive proofs are usually hard to follow; it's easy to get lost in the details, losing track of what n, k, and k+1 are supposed to be.
For other explanations and examples of mathematical induction, see
Proof by Mathematical Induction
Sum of n Odd Numbers
Inductive Misunderstanding | {
"domain": "themathdoctors.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9854964194566753,
"lm_q1q2_score": 0.8596952109972292,
"lm_q2_score": 0.8723473713594992,
"openwebmath_perplexity": 641.1596823095123,
"openwebmath_score": 0.7628352046012878,
"tags": null,
"url": "https://www.themathdoctors.org/more-on-0-999/"
} |
Proof by Mathematical Induction
Sum of n Odd Numbers
Inductive Misunderstanding
Here is the proof. As he’ll explain, rather than taking n as 1, 2, 3, …, he is taking n as the exponent, -1, -2, -3, …, going in the negative direction and so reversing the usual procedure.
In our case, we want to show that when multiplying the nth digit by 10, we get a 9 in the (n+1)st digit, no carry to the (n+2)nd digit, and a 0 in the nth digit (counting from the right.) We want no carry to the (n+2)nd digit so that we don't have to adjust the previous (greater place value) digits multiplied because of a "retroactive carry." We want a 0 in the nth digit so that it will not produce further carries when multiplying the next digit (smaller place value) by 10. Mathematically:
10 * (9*10^n) = 9 * 10^(n+1)
We want to show that:
a) the nth digit is 0
b) the (n+1)st digit is 9
c) the (n+2)nd digit is 0
To minimize the confusion, I'm going to use n = -1 as my basis step, and show that if it is true for n = k, then it is true for n = k-1 as my inductive step. This is the "negative" of the standard, but it will eliminate the need for using -k's and -(k+1)'s, etc. in my equations.
Step 1: Show that it is the case for n = -1
10 * [9*10^(-1)] = 9 * 10^(-1+1)
10 * (9*0.1) = 9*10^0
10 * 0.9 = 9*1
9 = 9
a) the -1st digit (the tenths place) is 0: True
b) the 0th digit (the units place) is 9: True
c) the 1st digit (the tens place) is 0: True
So we've proved the basis. Now for the inductive step.
Step 2: Prove that if it's true for n = k, then it's true for n = k-1.
Let's let d = the initial (k+1)st digit. This digit is 0 from conclusion (a) of the previous step. Then:
10 * (9*10^k) + d = 9 * 10^(k+1)
9 * (10*10^k) + 0 = 9 * 10^(k+1)
9 * 10^(k+1) = 9 * 10^(k+1)
a) the kth digit is 0: True - 9*10^(k+1) has a 0 in the kth place.
b) the (k+1)st digit is 9: True - 9*10^(k+1) has a 9 in the (k+1)st place
c) the (k+2)nd digit is 0: True - 9*10^(k+1) has a 9 in the (k+2)nd place | {
"domain": "themathdoctors.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9854964194566753,
"lm_q1q2_score": 0.8596952109972292,
"lm_q2_score": 0.8723473713594992,
"openwebmath_perplexity": 641.1596823095123,
"openwebmath_score": 0.7628352046012878,
"tags": null,
"url": "https://www.themathdoctors.org/more-on-0-999/"
} |
So, if 10 times the kth digit is 9*10^(k+1), then 10 times the (k-1)st digit is 9*10^k, and thus 10 times the (k-2)nd digit is 9*10^(k-1), and thus...
Therefore 10 * 0.999... = 9.999...
Q.E.D.
More could be done to make this foolproof:
Depending on the skepticism of the intended audience, you may have to use a similar inductive proof for the step:
9.999...
- 0.999...
--------
9.000...
For most audiences, the following will do:
Let x = 0.999...
then 9.999... = 9 + x
so 9.999... - 0.999... = (9 + x) - x = 9
However, some audiences (like your teacher, perhaps) will not be satisfied that algebraic operations work on "infinite strings" (i.e. they won't be convinced that x - x = 0 when x is an infinite repeating decimal.)
I think some of the other explanations discussed last time may be more useful for certain kinds of skeptics; but it’s good to have a variety of approaches, to meet a variety of objections.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | {
"domain": "themathdoctors.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9854964194566753,
"lm_q1q2_score": 0.8596952109972292,
"lm_q2_score": 0.8723473713594992,
"openwebmath_perplexity": 641.1596823095123,
"openwebmath_score": 0.7628352046012878,
"tags": null,
"url": "https://www.themathdoctors.org/more-on-0-999/"
} |
# value of $\int_{-\infty}^{\infty}\arcsin\frac1{\cosh x}\,dx$
I want to know the value of $$I=\int_{-\infty}^{\infty}\arcsin\frac1{\cosh x}\,dx$$ The Symbolab integral calculator says that the integral diverges, but when one graphs it obvious that it converges. So what is the value?
I was thinking that I might try Feynman integration, but I can't think of the right substitution.
From the answer provided by @user10354138, we can reach $$\int\arcsin\frac1{\cosh x}dx=i\operatorname{Li}_2(i\phi)-i\operatorname{Li}_2(-i\phi)+C$$ Where $$\phi=\tan\bigg(\frac12\arcsin\frac1{\cosh x}\bigg)$$ And $$\operatorname{Li}_2(z)=\sum_{n\geq1}\frac{z^n}{n^2}$$ is the Di-logarithm.
Wolfy says it is 4 times the Catalan's constant.
One (not optimal) way to derive this is \def\sech{\operatorname{sech}} \begin{align*} \int_{-\infty}^\infty\arcsin\sech x\,\mathrm{d}x&=2\int_0^\infty\arcsin\sech x\,\mathrm{d}x\\ &=2\int_0^1\frac{\arcsin u\,\mathrm{d}u}{u\sqrt{1-u^2}}\quad(u=\sech x)\\ &=2\int_0^{\pi/2}\frac{\theta\,\mathrm{d}\theta}{\sin\theta}\quad(u=\sin\theta)\\ &=2\int_0^1\frac{2\tan^{-1}t\,\frac{2\,\mathrm{d}t}{1+t^2}}{\frac{2t}{1+t^2}}\quad(t=\tan\tfrac12\theta)\\ &=4\int_0^1\frac{\tan^{-1}t}{t}\,\mathrm{d}t\\ &=4\int_0^1\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}t^{2n}\,\mathrm{d}t\\ &=4\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}=4G\\ \end{align*}
• Dude that's dope! Thank you! – clathratus Oct 31 '18 at 3:31
• This substitution simplifies to $t=e^{-x}$. – J.G. Oct 31 '18 at 8:44
• how do you get from $\frac{\theta d\theta}{\sin\theta}$ to $$\frac{2\arctan(t)\frac{2dt}{1+t^2}}{\frac{2t}{1+t^2}}$$ With the substitution $t=\tan\frac\theta2$? – clathratus Nov 1 '18 at 23:06
• I keep getting $$\frac{\theta d\theta}{\sin\theta}=\frac{2\arctan t}{t\sqrt{t^2+1}}dt$$ – clathratus Nov 1 '18 at 23:10
• @clathratus This is the universal trigonometric substitution – user10354138 Nov 1 '18 at 23:25 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9854964203086181,
"lm_q1q2_score": 0.8596952101051306,
"lm_q2_score": 0.872347369700144,
"openwebmath_perplexity": 708.5624130677423,
"openwebmath_score": 0.9996808767318726,
"tags": null,
"url": "https://math.stackexchange.com/questions/2978611/value-of-int-infty-infty-arcsin-frac1-cosh-x-dx"
} |
Alternatively, you can integrate by parts: \begin{align} \int \limits_{-\infty}^\infty \arcsin(\operatorname{sech}(x)) \, \mathrm{d} x &= 2 \int \limits_0^\infty \arcsin(\operatorname{sech}(x)) \, \mathrm{d} x \\ &= 2x \arcsin(\operatorname{sech}(x)) \Bigg \rvert_{x=0}^{x=\infty} - 2 \int \limits_0^\infty x \frac{- \sinh(x) \operatorname{sech}^2(x)}{\sqrt{1-\operatorname{sech}^2(x)}} \, \mathrm{d} x \\ &= 2 \int \limits_0^\infty \frac{x}{\cosh(x)} \, \mathrm{d} x = 4 \sum \limits_{n=0}^\infty (-1)^n \int \limits_0^\infty x \, \mathrm{e}^{-(2n+1) x} \, \mathrm{d} x \\ &= 4 \Gamma(2) \sum \limits_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} = 4 \mathrm{G} \, . \end{align}
• I'm a fan of the alternative approach. Thank you! – clathratus Oct 31 '18 at 14:59
Yet another alternative approach: once shown that $$\int_{\mathbb{R}}\frac{dx}{\cosh(x)^{2k+1}} = \frac{\pi \binom{2k}{k}}{4^k}\tag{1}$$ and recalled that $$\arcsin z = \sum_{k\geq 0}\frac{\binom{2k}{k}}{4^k(2k+1)}z^{2k+1} \tag{2}$$ we have the following identity: $$\int_{\mathbb{R}}\arcsin\frac{1}{\cosh x}\,dx = \pi\sum_{k\geq 0}\frac{1}{2k+1}\left[\frac{1}{4^k}\binom{2k}{k}\right]^2.\tag{3}$$ Now we may invoke a function whose Maclaurin series involves squared central binomial coefficients, namely the complete elliptic integral of the first kind $$K(x)$$, here denoted according to Mathematica's notation (the argument of $$K$$ is the elliptic modulus): $$\sum_{k\geq 0}\left[\frac{1}{4^k}\binom{2k}{k}\right]^2 x^{2k}=\frac{2}{\pi}K(x^2)\tag{4}$$ leading to: $$\int_{\mathbb{R}}\arcsin\frac{1}{\cosh x}\,dx = 2\int_{0}^{1} K(x^2)\,dx = \int_{0}^{1}\frac{K(x)}{\sqrt{x}}\,dx.\tag{5}$$ At last, we recall that both $$K(x)$$ and $$\frac{1}{\sqrt{x}}$$ have fairly simple Fourier-Legendre series expansions: $$K(x)=2\sum_{n\geq 0}\frac{P_n(2x-1)}{2n+1},\qquad \frac{1}{\sqrt{x}}=2\sum_{n\geq 0}(-1)^n P_n(2x-1)$$ and by the orthogonality of shifted Legendre polynomials | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9854964203086181,
"lm_q1q2_score": 0.8596952101051306,
"lm_q2_score": 0.872347369700144,
"openwebmath_perplexity": 708.5624130677423,
"openwebmath_score": 0.9996808767318726,
"tags": null,
"url": "https://math.stackexchange.com/questions/2978611/value-of-int-infty-infty-arcsin-frac1-cosh-x-dx"
} |
$$\int_{\mathbb{R}}\arcsin\frac{1}{\cosh x}\,dx =\int_{0}^{1}\frac{K(x)}{\sqrt{x}}\,dx = 4\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2} = 4G.\tag{6}$$
• This is really cool. I'll have to look at this one a little more in order to understand it better... But this is great. Thanks! – clathratus Oct 31 '18 at 18:37
• You seem to link any special function with any other special function. The link between elliptic integral and Legendre function is pretty smart. +1 – Paramanand Singh Nov 1 '18 at 16:41
• How do we show $$\int_{\Bbb R}\frac{\mathrm dx}{\cosh(x)^{2k+1}}=\frac{\pi{2k\choose k}}{4^k}$$ It looks a lot like a Beta integral – clathratus Dec 21 '18 at 0:56
• @clathratus: it is, indeed, it is enough to set $\frac{1}{\cosh(x)}=u$. – Jack D'Aurizio Dec 21 '18 at 10:48 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9854964203086181,
"lm_q1q2_score": 0.8596952101051306,
"lm_q2_score": 0.872347369700144,
"openwebmath_perplexity": 708.5624130677423,
"openwebmath_score": 0.9996808767318726,
"tags": null,
"url": "https://math.stackexchange.com/questions/2978611/value-of-int-infty-infty-arcsin-frac1-cosh-x-dx"
} |
# Applying Gauss' Divergence Theorem to given integral
We are given the following two integrals:
$$\iint\limits_S D_n f\:dS$$ and $$\iiint\limits_B \nabla \cdot (\nabla f) \: dV$$
where $$S$$ is the portion of the sphere $$x^2+y^2+z^2=a^2$$ in the first octant, $$n$$ is the unit normal vector to $$S$$ at $$(x,y,z)$$ and $$f(x,y,z) = ln(x^2+y^2+z^2)$$
For the first integral, we have: $$D_n f = \nabla f \cdot n$$ and $$n = \frac{1}{a}$$
Since $$\nabla f = \: <\frac{2x}{x^2+y^2+z^2}, \frac{2y}{x^2+y^2+z^2}, \frac{2z}{x^2+y^2+z^2}>$$, this gives $$D_n f = \nabla f \cdot n = \frac{2}{a}$$.
So the integral becomes
$$\iint\limits_S D_nf \: dS = \iint\limits_S \frac{2}{a} \: dS = \frac{2}{a} \iint\limits_S dS = \frac{2}{a} \cdot A(S) = \frac{2}{a} \cdot 4\pi a^2 \cdot \frac{1}{8} = \pi a$$ ($$\frac{1}{8}$$ since the sphere is in the first octant)
For the second integral, we have $$\nabla \cdot (\nabla f) = \frac{2}{a^2}$$ (since $$x^2 + y^2 + z^2 = a^2$$). So the integral becomes $$\iiint\limits_B \nabla \cdot (\nabla f) \: dV = \frac{2}{a^2} \iiint\limits_B dV = \frac{2}{a^2} \cdot V(B) = \frac{2}{a^2} \cdot \frac{4}{3} \pi a^3 \cdot \frac{1}{8} = \frac{1}{3} \pi a$$
However, according to Gauss' Divergence theorem, these two integrals should be equal to each other right? Since we have $$\iiint\limits_B \nabla \cdot (\nabla f) \: dV = \iint\limits_S \nabla f \cdot dS = \iint\limits_S \nabla f \cdot n \: dS = \iint\limits_S D_n f \: dS$$
So how is it possible that I get two different answers, even though the Divergence theorem shows that the two integrals should be the same? | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.985496419030704,
"lm_q1q2_score": 0.8596952089903457,
"lm_q2_score": 0.8723473697001441,
"openwebmath_perplexity": 208.79121913026995,
"openwebmath_score": 0.9417594075202942,
"tags": null,
"url": "https://math.stackexchange.com/questions/4081507/applying-gauss-divergence-theorem-to-given-integral"
} |
• You're not counting the flat parts of the surface which are quarter circles in the coordinate planes. – B. Goddard Mar 29 at 11:09
• Oh I see, so if I understand correctly, is it true that the results of the first two integrals are correct, but the application of Gauss' theorem isn't? Because the surface S is not the same as the boundary surface of B, since the boundary surface contains extra flat parts? – Stallmp Mar 29 at 11:14
To apply divergence theorem, you must have a closed surface. So we close the surface by placing $$3$$ quarter disks in plane $$x = 0, y = 0, z = 0$$.
Please note that when you are doing volume integral, you cannot equate $$\nabla \cdot (\nabla f) = \displaystyle \frac{2}{x^2+y^2+z^2} = \frac{2}{a^2}$$. It should rather be,
$$\nabla \cdot (\nabla f) = \displaystyle \frac{2}{x^2+y^2+z^2} = \frac{2}{\rho^2}$$
So the integral becomes,
$$\displaystyle \int_0^{\pi/2} \int_0^{\pi/2} \int_0^a \frac{2}{\rho^2} \ \cdot\rho^2 \cdot \sin \phi \ d\rho \ d\phi \ d\theta = \pi a$$.
Now to find flux through $$S$$, we must subtract flux through planar surfaces $$x = 0, y = 0, z = 0$$ but in this case, they are simply zero. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.985496419030704,
"lm_q1q2_score": 0.8596952089903457,
"lm_q2_score": 0.8723473697001441,
"openwebmath_perplexity": 208.79121913026995,
"openwebmath_score": 0.9417594075202942,
"tags": null,
"url": "https://math.stackexchange.com/questions/4081507/applying-gauss-divergence-theorem-to-given-integral"
} |
• simply because $x^2+y^2+z^2 = a^2$ only on the surface of the sphere, not inside it. We are talking volume integral which includes all points inside the sphere too and so we write $x^2+y^2+z^2 \leq a^2$. – Math Lover Mar 29 at 11:31
• Yes you are right about surface $S$. What you get from divergence theorem will include flux through closed surface which is spherical surface $S$ and $3$ planar surfaces as mentioned in my answer. We are supposed to calculate that separately and subtract from the result that we get from divergence theorem to obtain flux only through spherical surface $S$. But in this case they are zero. Do you see why or do you need an explanation? – Math Lover Mar 29 at 11:39
• Yes this is indeed what I thought, thank you very much for the explanation! I see why they are zero as well when calculating the flux through the planar surfaces. – Stallmp Mar 29 at 11:45
• Yes you are right, we need to show they sum to zero. Note that $\iint\limits_{S_1} D_n f \: dS = \iint\limits_{S_2} D_n f \: dS = \iint\limits_{S_3} D_n f \: dS = 0$ – Math Lover Mar 29 at 12:34
• – Math Lover Mar 29 at 12:35 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.985496419030704,
"lm_q1q2_score": 0.8596952089903457,
"lm_q2_score": 0.8723473697001441,
"openwebmath_perplexity": 208.79121913026995,
"openwebmath_score": 0.9417594075202942,
"tags": null,
"url": "https://math.stackexchange.com/questions/4081507/applying-gauss-divergence-theorem-to-given-integral"
} |
# Method for Extracting the $n$th root of a number?
I was wondering if there exists a reliable way to extract the $$n$$th root of a given number. For example, if you have a large perfect square such as 10,404, how would I go about taking its square root by hand?
And what about cube roots? Is there a way to find the cube root of any number by hand?
If there is a way to take the cube and square roots of any number, is there a technique that can be applied to extract the cube or square root of polynomials? What about any root?
I think that, if there is a way to take the $$n$$th root of any number of polynomial, I would benefit from learning it, since I consider being able to work with roots important to my algebra foundation.
Thanks
• For the square root, there is a method similar to the usual division by a number. If we know that the number is a perfect square we can restrict the possibilities (ending digits , residue mod 9). For cube roots and higher , I am not aware of an efficient method by hand. Feb 25 at 18:58
• For square roots we can use the algorithm described here under "Digit-by-digit calculation" / "Decimal (base 10)" Feb 25 at 19:03
• @StefanOctavian I've seen similar algorithms for higher roots done (somehow you take 3-digits at a time for cube roots.) But they stopped teaching the square-root extraction the year before I would have had to learn it. Feb 25 at 20:38
If you would like to compute $$\sqrt[p]{a}$$, the following iterative process will do it to whatever accuracy you please:
First, make a guess and call it $$x_0$$. Then perform this iteration:
$$x_{n+1} = \frac{(p-1)x_n^p+a}{px^{k-1}}.$$
For instance, if you want $$\sqrt[5]{11}$$, guess that $$x_0=2$$. The iteration looks like
$$x_{n+1} = \frac{4x_n^5+11}{5x_n^4}.$$
So we have $$x_1 = \frac{4\cdot2^5+11}{5\cdot 2^4} = \frac{139}{80}=1.7375.$$
$$x_2=\frac{4\cdot1.7375^5+11}{5\cdot 1.7375^4} = 1.631392308.$$
This number to the 5th power is $$11.55558806$$, so we're getting close. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9854964186047326,
"lm_q1q2_score": 0.8596952020775964,
"lm_q2_score": 0.8723473630627235,
"openwebmath_perplexity": 259.7634815264026,
"openwebmath_score": 0.8494479060173035,
"tags": null,
"url": "https://math.stackexchange.com/questions/4391113/method-for-extracting-the-nth-root-of-a-number"
} |
This number to the 5th power is $$11.55558806$$, so we're getting close.
$$x_3=\frac{4\cdot1.631392308^5+11}{5\cdot 1.631392308^4} = 1.615704970.$$
And $$1.615704970^5 = 11.0105\ldots.$$
This is an implementation of Newton's Method on the polynomial $$x^p-a$$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9854964186047326,
"lm_q1q2_score": 0.8596952020775964,
"lm_q2_score": 0.8723473630627235,
"openwebmath_perplexity": 259.7634815264026,
"openwebmath_score": 0.8494479060173035,
"tags": null,
"url": "https://math.stackexchange.com/questions/4391113/method-for-extracting-the-nth-root-of-a-number"
} |
Related Rates:
Gas is escaping from a spherical balloon at the rate of 2ft^3/min. How fast is the surface area shrinking (ds/dt) when the radius is 12ft? (A sphere of radius r has volume v=4/3 pi r^3 and surface area S=4pi r^2.)
Remember that ds/dt = ds/dr X dr/dt
Step 1: Find ds/dr
I have no clue where to start. I was going to set 2 equal to 4 pi r^2 and then take the derivative but I really have no idea.
Step 2: Find dr/dr. (HInt dv/dt= dv/dr X dr/dt
Would I set 2 equal to volume and then take derivative?
Step 3: Find ds/dt?
Step 4: Evaluate ds/dt when the radius is 12ft.
After I find my equation I would just plug in 12 correct?
1. 👍
2. 👎
3. 👁
1. S = 4pi*r^2
dS/dr = 4pi*2r = 8pi*r
V = (4/3)pi*r^3
dV/dr = 4pi*r^2
dV/dt = dV/dr * dr/dt
2 = 4pi*r^2 * dr/dt
dr/dt = 1/(2pi*r^2)
dS/dt = dS/dr * dr/dt
= 8pi*r * 1/(2pi*r^2)
= 4/r
If r = 12, just plug the value to the last equation
1. 👍
2. 👎
2. How did you get 12?
1. 👍
2. 👎
3. Nevermind I meant how did you get 4/r but I figured it out. Thank You!!
1. 👍
2. 👎
4. Is this correct for the last part?
(8)(3.14)(12) X (1/(2)(3.14)(12^2) = 4/12
1. 👍
2. 👎
5. Yes
1. 👍
2. 👎
6. Ok do I actually have to do the calculations though?
1. 👍
2. 👎
7. If you are not given the value, that means you have to write the steps I told you until the very last equation.
If you are given the value (of r), you don't have to put the value in every equation. Just follow my steps until the last equation, then plug the value after that. It's easier that way.
1. 👍
2. 👎
8. OK thank you!
1. 👍
2. 👎
9. Simplify the answer (4/12) into 1/3
1. 👍
2. 👎
## Similar Questions
1. ### calculas
i didnot under stand the question ? need help Gas is escaping a spherical balloon at the rate of 4 cm3 per minute. How fast is the surface area of the balloon shrinking when the radius of the balloon is 24 cm? Given volume of
2. ### Math | {
"domain": "jiskha.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9706877692486438,
"lm_q1q2_score": 0.8596716455572566,
"lm_q2_score": 0.8856314798554445,
"openwebmath_perplexity": 786.0660610559516,
"openwebmath_score": 0.8956266641616821,
"tags": null,
"url": "https://www.jiskha.com/questions/509900/related-rates-gas-is-escaping-from-a-spherical-balloon-at-the-rate-of-2ft-3-min-how-fast"
} |
2. ### Math
A spherical balloon is to be deflated so that its radius decreases at a constant rate of 12 cm/min. At what rate must air be removed when the radius is 5 cm? Must be accurate to the 5th decimal place. It seems like an easy
3. ### Word Problem-Calculus
A spherical balloon is being inflated in such a way that its radius increases at a rate of 3 cm/min. If the volume of the balloon is 0 at time 0, at what rate is the volume increasing after 5 minutes? my answer is 45 cm/min. is
4. ### Calculus
a spherical balloon is inflated with gas at the rate of 500 cubic centimeters per minute. how fast is the radius of the balloon increasing at the instant the radius is 30 centimeters?
1. ### Calculus
A spherical balloon is inflated at a rate of 10 cubic feet per minute. How fast is the radius of the balloon changing at the instant the radius is 4 feet? And The radius of a circle is decreasing at a rate of 2 ft/minute. Find the
2. ### help on calculus
A spherical balloon is being filled with air in such a way that its radius is increasing at a rate of 2 centimeters per second. At what rate is the volume of the balloon changing at the instant when its surface has an area of 4 pi
3. ### Calculus
Air is being pumped into a spherical balloon so that its volume increases at a rate of 30 cubic centimeters per second. How fast is the surface area of the balloon increasing when its radius is 19 cm? V= 4/3*pi*r^3 S= 4 pi r^2
4. ### calculus help thanks!
The gas law for an ideal gas at absolute temperature T (in kelvins), pressure P (in atmospheres), and volume V (in liters) is PV = nRT, where n is the number of moles of the gas and R = 0.0821 is the gas constant. Suppose that, at
1. ### Math
A spherical balloon is inflated with helium at the rate of 100pi ft ^3/min. How fast is the ballon's radius increasing at the instant the radius is 5ft.?
2. ### Mathematics | {
"domain": "jiskha.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9706877692486438,
"lm_q1q2_score": 0.8596716455572566,
"lm_q2_score": 0.8856314798554445,
"openwebmath_perplexity": 786.0660610559516,
"openwebmath_score": 0.8956266641616821,
"tags": null,
"url": "https://www.jiskha.com/questions/509900/related-rates-gas-is-escaping-from-a-spherical-balloon-at-the-rate-of-2ft-3-min-how-fast"
} |
2. ### Mathematics
A spherical balloon is blown up so that it's volume increases constantly at the rate of 2cm^3 per second .find the rate of the increase of the radius when the volume of the balloon is 50cm^3
3. ### defferential calculus
A balloon is rising vertically 150 m from an observer. At exactly 1 min, the angle of elevation is 29 deg 28 min. How fast is the balloon rising at that instant?
4. ### Calculus
A spherical balloon is being inflated and the radius of the balloon is increasing at a rate of 2cm/min. At what rates are the volume and surface area of the balloon increasing when the radius is 5cm? | {
"domain": "jiskha.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9706877692486438,
"lm_q1q2_score": 0.8596716455572566,
"lm_q2_score": 0.8856314798554445,
"openwebmath_perplexity": 786.0660610559516,
"openwebmath_score": 0.8956266641616821,
"tags": null,
"url": "https://www.jiskha.com/questions/509900/related-rates-gas-is-escaping-from-a-spherical-balloon-at-the-rate-of-2ft-3-min-how-fast"
} |
# Relation between $\bigcap_{i \in I}A_i$ and $\bigcap_{i=1}^{n}A_i$
Let I be a nonempty set and a family of sets such that every element of the family is a subset of U.
$\mathcal F = \{A_i | i \in I\}$
I understand the meaning of this operation:
$$\bigcap_{i \in I}A_i$$
That's the intersection of all the elements of the sets of the family $\mathcal F$.
But I don't truly understand what this other operation means and what's the relation between the above one. $$\bigcap_{i=1}^{n}A_i$$
I understand that this operation is also an intersection but what I am trying to understand is the relation between $\bigcap_{i \in I}A_i$ and $\bigcap_{i=1}^{n}A_i$
$$\bigcap_{i=1}^{n}A_i \subseteq\bigcap_{i \in I}A_i$$
Is the relation above true?
• In the second case we have $I = \{ 1,2,\ldots, n \}$. – Mauro ALLEGRANZA Jun 23 '18 at 15:28 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9706877717925422,
"lm_q1q2_score": 0.8596716316944107,
"lm_q2_score": 0.8856314632529871,
"openwebmath_perplexity": 211.5068487325436,
"openwebmath_score": 0.9462869763374329,
"tags": null,
"url": "https://math.stackexchange.com/questions/2829529/relation-between-bigcap-i-in-ia-i-and-bigcap-i-1na-i"
} |
• In the second case we have $I = \{ 1,2,\ldots, n \}$. – Mauro ALLEGRANZA Jun 23 '18 at 15:28
The notation $$\bigcap _{i=1}^nA_i$$ denotes the same as $$\bigcap_{i\in I}A_i$$ for the special case $I=\{1,2,\ldots, n\}$.
• So $\bigcap_{i\in I}A_i$ is the general case and $\bigcap _{i=1}^nA_i$ is a specific case for which we give concrete values to i? – adriana634 Jun 23 '18 at 15:40
• @adriana634 Yes. One situation where we see both notations used simultaneously is when $I=\mathbb{N}$; then "$\bigcap_{i=1}^nA_i$" means "$\bigcap_{i\in \{1,2,..., n\}}A_i$," which is generally a superset of $\bigcap_{i\in I}A_i$. – Noah Schweber Jun 23 '18 at 15:42
• @Noah Schweber But for example $\bigcap_{i=1}^{5}A_i$ means $I=\{1,2,\ldots, 5\}$, in this example $\bigcap_{i\in I}A_i$ is a superset of $\bigcap_{i=1}^{5}A_i$? I don't understand why you said $\bigcap_{i=1}^nA_i$ is a superset of $\bigcap_{i\in I}A_i$. – adriana634 Jun 23 '18 at 16:00
• @adriana634 No. The intersection over a bigger index set yields a smaller set. E.g. if we take $I=\mathbb{N}$ and $A_i=\{i, i+1, i+2, ...\}$ for $i\in I$, then $\bigcap_{i=1}^5A_i=\{5, 6,7,...\}$ but $\bigcap_{i\in I}A_i=\emptyset$. More abstractly: $$I\supseteq J\implies \bigcap_{i\in I}A_i\subseteq \bigcap_{i\in J}A_i.$$ – Noah Schweber Jun 23 '18 at 16:02 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9706877717925422,
"lm_q1q2_score": 0.8596716316944107,
"lm_q2_score": 0.8856314632529871,
"openwebmath_perplexity": 211.5068487325436,
"openwebmath_score": 0.9462869763374329,
"tags": null,
"url": "https://math.stackexchange.com/questions/2829529/relation-between-bigcap-i-in-ia-i-and-bigcap-i-1na-i"
} |
GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 21 Nov 2018, 02:01
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
## Events & Promotions
###### Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day!
November 22, 2018
November 22, 2018
10:00 PM PST
11:00 PM PST
Mark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)
• ### Free lesson on number properties
November 23, 2018
November 23, 2018
10:00 PM PST
11:00 PM PST
Practice the one most important Quant section - Integer properties, and rapidly improve your skills.
# In the addition shown above, A, B, C, and D represent
Author Message
TAGS:
### Hide Tags
Manager
Joined: 09 Nov 2012
Posts: 63
In the addition shown above, A, B, C, and D represent [#permalink]
### Show Tags
15 Oct 2013, 07:17
7
41
00:00
Difficulty:
95% (hard)
Question Stats:
42% (03:00) correct 58% (02:47) wrong based on 776 sessions
### HideShow timer Statistics
ABC
+BCB
CDD
In the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ? | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637612961505,
"lm_q2_score": 0.8596637612961505,
"openwebmath_perplexity": 1315.5501724991552,
"openwebmath_score": 0.7710837125778198,
"tags": null,
"url": "https://gmatclub.com/forum/in-the-addition-shown-above-a-b-c-and-d-represent-161656.html"
} |
A. 8
B. 10
C. 12
D. 14
E. 18
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4488
Re: In the addition shown above, A, B, C, and D represent [#permalink]
### Show Tags
15 Oct 2013, 10:47
17
12
saintforlife wrote:
ABC
+BCB
CDD
In the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?
A. 8
B. 10
C. 12
D. 14
E. 18
I'm happy to help with this.
First, look at the one's column. C + B produces a one's unit of D, and we don't know whether anything carries to the ten's place.
But now, look at the ten's place --- B + C again produces a digit of D --- that tells us definitively that nothing carried, and that B + C = D. B & C have a single digit number as a sum.
Now, C = A + B, because again, in the hundreds column, nothing carries here.
Notice that, since C = A + B and D = B + C, B must be smaller than both B and C. We want to make the product of A & B large, so we want to make those individual digits large. Well, if we make B large, then that makes C large, and then B + C would quickly become more than a one-digit sum, which is not allowed. Think about it this way. Let's just assume D = 9, the maximum value.
D = 9 = B + C = B + (A + B) = A + 2B
We want to pick A & B such that A + 2B = 9 and A*B is a maximum. It makes sense that B would be smaller.
Try A = 7, B = 1. Then A + 2B = 9 and A*B = 7
Try A = 5, B = 2. Then A + 2B = 9 and A*B = 10
Try A = 3, B = 3. Then A + 2B = 9 and A*B = 9
Try A = 1, B = 4. Then A + 2B = 9 and A*B = 4
Indeed, as B gets bigger, the product gets less. This seems to imply that the biggest possible product is 10. This corresponds to A = 5, B = 2, C = 7, and D = 9, and the original addition problem becomes
527
+272
799
Thus, the maximum product is 10, and answer = (B).
Does all this make sense?
Mike
_________________
Mike McGarry
Magoosh Test Prep | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637612961505,
"lm_q2_score": 0.8596637612961505,
"openwebmath_perplexity": 1315.5501724991552,
"openwebmath_score": 0.7710837125778198,
"tags": null,
"url": "https://gmatclub.com/forum/in-the-addition-shown-above-a-b-c-and-d-represent-161656.html"
} |
Does all this make sense?
Mike
_________________
Mike McGarry
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Current Student
Joined: 31 Mar 2013
Posts: 66
Location: India
GPA: 3.02
Re: In the addition shown above, A, B, C, and D represent [#permalink]
### Show Tags
06 Dec 2013, 10:19
8
1
saintforlife wrote:
How do we do all that in less than 2 mins, is my only question :)
Posted from my mobile device
This problem can be solved much faster if you start with the options. As Mike has wonderfully explained, none of the digits will have a carry over.
Option E is $$18 = 9 * 2$$(no other case is possible). Now$$9+2>=10$$. Therefore, it will have a carry over digit 1. So reject this option.
Option D is $$14 = 7 * 2$$(no other case is possible). Since $$C=A+B$$, therefore $$C= 9$$. Since $$C=9$$ and $$B$$ is a non zero digit, $$C+B>=10$$. Therefore, it will have a carry over digit 1. So reject this option.
Option C is 12. This will have 2 cases-
$$Case 1--> 12 = 4 * 3$$
Here $$C= 4+3=7$$. Now $$7+4>=10$$ and [$$7+3>=10$$. Therefore, both cases will have a carry over digit 1. So reject this option.
(OR) $$Case 2--> 6 * 2$$.
Here $$C= 6+2=8$$. Now $$8+2>=10$$ and $$8+6>=10$$. Therefore, both cases will have a carry over digit 1. So reject this option.
Option B is $$10 = 5 * 2$$. If you follow the same set of steps you'll notice that there will not be any carry over digit for the case $$C=7, A=5, B=2$$. So right answer!
##### General Discussion
Manager
Joined: 09 Nov 2012
Posts: 63
Re: In the addition shown above, A, B, C, and D represent [#permalink]
### Show Tags
23 Oct 2013, 15:44
2
How do we do all that in less than 2 mins, is my only question
Posted from my mobile device
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4488
Re: In the addition shown above, A, B, C, and D represent [#permalink]
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637612961505,
"lm_q2_score": 0.8596637612961505,
"openwebmath_perplexity": 1315.5501724991552,
"openwebmath_score": 0.7710837125778198,
"tags": null,
"url": "https://gmatclub.com/forum/in-the-addition-shown-above-a-b-c-and-d-represent-161656.html"
} |
### Show Tags
23 Oct 2013, 15:54
6
saintforlife wrote:
How do we do all that in less than 2 mins, is my only question
Dear Saint For Life,
Problems such as this are quite out-of-the-box. To some extent, the GMAT intends us to handle many of the other questions in 90 secs or less, so that we have a bit of a time-cushion when we run into one of these oddball questions.
Having said that, the more familiar you are with number properties, the faster it will go. In this problem, it took me quite some time to write out everything in verbal form, but I saw things relatively quickly. As you practice seeing patterns, you will see them more quickly, even if they are hard to explain to someone else. You may find this post helpful:
http://magoosh.com/gmat/2013/how-to-do- ... th-faster/
Mike
_________________
Mike McGarry
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Math Expert
Joined: 02 Sep 2009
Posts: 50727
Re: In the addition shown above, A, B, C, and D represent [#permalink]
### Show Tags
23 Oct 2013, 23:13
5
2
saintforlife wrote:
ABC
+BCB
CDD
In the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?
A. 8
B. 10
C. 12
D. 14
E. 18
Similar questions to practice:
tough-tricky-set-of-problems-85211.html#p638336
Hope it helps
_________________
Intern
Joined: 16 Jun 2013
Posts: 14
GMAT 1: 540 Q34 V30
GMAT 2: 700 Q43 V42
Re: In the addition shown above, A, B, C, and D represent the [#permalink]
### Show Tags
26 Oct 2013, 21:47
2
... got E as an answer. However, it took me 6 minutes, since I plugged in the answer options.
Is there any faster way, or am I just too slow in back-solving?
TirthankarP wrote: | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637612961505,
"lm_q2_score": 0.8596637612961505,
"openwebmath_perplexity": 1315.5501724991552,
"openwebmath_score": 0.7710837125778198,
"tags": null,
"url": "https://gmatclub.com/forum/in-the-addition-shown-above-a-b-c-and-d-represent-161656.html"
} |
TirthankarP wrote:
In the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?
A) 8
B) 10
C) 12
D) 14
E) 18
My explanation:
According to the question stem C+B=D and D<10, since the second addition (2nd column) is C+B=D again (if there were a carry, the second column's result would not be D again), and A+B=C
Now I break up the answer options:
A) A*B=8, so A and B -> 2 and 4 or 1 and 8. A+B=2+4=6=C, C+B=6+2=8=D, which is smaller than 10, thus the answer is not "out", but there could be a larger possible value of A*B
B) A*B=10, so A and B -> 2 and 5. A+B=2+5=7=C, C+B=7+2=9=D, which is smaller than 10, thus the is also possible. Let's look for the next options.
For C-E:
C) A*B=12, so A and B -> 6 and or 4 and 3.
D) A*B=14, so A and -> 7 and 2.
E) A*B=18, so A and B -> 9 and 2 or 3 and 6.
Do the same process with those numbers and you will find that all will yield a sum of C+B>10, thus the constraint of D<10 is not satisfied. The answer options are not possible.
Because 10>8 answer option B) is the largest possible.
Manager
Joined: 29 Apr 2013
Posts: 94
Location: India
Concentration: General Management, Strategy
GMAT Date: 11-06-2013
WE: Programming (Telecommunications)
Re: In the addition shown above, A, B, C, and D represent the [#permalink]
### Show Tags
27 Oct 2013, 00:04
Chiranjeevee wrote:
TirthankarP wrote:
Attachment:
In the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?
A) 8
B) 10
C) 12
D) 14
E) 18
I do a thorough search before posting any new question. But this time I couldn't find this link
Forum moderator may delete this thread
_________________
Do not forget to hit the Kudos button on your left if you find my post helpful
Collection of some good questions on Number System | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637612961505,
"lm_q2_score": 0.8596637612961505,
"openwebmath_perplexity": 1315.5501724991552,
"openwebmath_score": 0.7710837125778198,
"tags": null,
"url": "https://gmatclub.com/forum/in-the-addition-shown-above-a-b-c-and-d-represent-161656.html"
} |
Collection of some good questions on Number System
Retired Moderator
Joined: 18 Sep 2014
Posts: 1127
Location: India
In the addition shown above, A, B, C, and D represent [#permalink]
### Show Tags
01 Jul 2015, 07:02
HKHR wrote:
saintforlife wrote:
How do we do all that in less than 2 mins, is my only question :)
Posted from my mobile device
This problem can be solved much faster if you start with the options. As Mike has wonderfully explained, none of the digits will have a carry over.
Option E is $$18 = 9 * 2$$(no other case is possible). Now$$9+2>=10$$. Therefore, it will have a carry over digit 1. So reject this option.
Option D is $$14 = 7 * 2$$(no other case is possible). Since $$C=A+B$$, therefore $$C= 9$$. Since $$C=9$$ and $$B$$ is a non zero digit, $$C+B>=10$$. Therefore, it will have a carry over digit 1. So reject this option.
Option C is 12. This will have 2 cases-
$$Case 1--> 12 = 4 * 3$$
Here $$C= 4+3=7$$. Now $$7+4>=10$$ and [$$7+3>=10$$. Therefore, both cases will have a carry over digit 1. So reject this option.
(OR) $$Case 2--> 6 * 2$$.
Here $$C= 6+2=8$$. Now $$8+2>=10$$ and $$8+6>=10$$. Therefore, both cases will have a carry over digit 1. So reject this option.
Option B is $$10 = 5 * 2$$. If you follow the same set of steps you'll notice that there will not be any carry over digit for the case $$C=7, A=5, B=2$$. So right answer!
Wonderful explanation. Just thought to add some more from my side. As we see only B satisfies all the necessary requirement(not to carry over any digits).
In option A, taking A=4 and B=2 also satisfies the condition as it is something like
426
+262
-------
688
But since we need maximum value well go for option B
Intern
Joined: 26 May 2017
Posts: 4
Re: In the addition shown above, A, B, C, and D represent [#permalink]
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637612961505,
"lm_q2_score": 0.8596637612961505,
"openwebmath_perplexity": 1315.5501724991552,
"openwebmath_score": 0.7710837125778198,
"tags": null,
"url": "https://gmatclub.com/forum/in-the-addition-shown-above-a-b-c-and-d-represent-161656.html"
} |
### Show Tags
30 Nov 2017, 18:03
Dear Mike, I was wondering why are we maximising D since it no where mentioned to do so. We have to maximise A*B right so if we choose C=1 and B=5 and A=4 i think we staisfy all conditions and our product of A*B (5*4) comes to be 20. Dont know where i am going wrong. Kindly help
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4488
Re: In the addition shown above, A, B, C, and D represent [#permalink]
### Show Tags
01 Dec 2017, 13:54
1
1
Sarthak.bhatt wrote:
Dear Mike, I was wondering why are we maximising D since it no where mentioned to do so. We have to maximise A*B right so if we choose C=1 and B=5 and A=4 i think we staisfy all conditions and our product of A*B (5*4) comes to be 20. Dont know where i am going wrong. Kindly help
Dear Sarthak.bhatt,
I'm happy to respond.
With all due respect, my friend, you are not interpreting the question correctly. There is absolutely no multiplication happening in this question. Here's the prompt again:
ABC
+BCB
CDD
In the addition shown above, A, B, C, and D represent
the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?
Thus, for example if C = 1, B = 5, and A = 4, then ABC would not be the product of those three numbers but instead the single three-digit number 451. With those numbers, the problem would be
451
+515
966
These choices satisfy the equation.
The problem is completely different from the way you were conceptualizing it, so the strategy is completely different from what it would be in the problem you had in mind.
Does this make sense?
Mike
_________________
Mike McGarry
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Intern
Joined: 21 Sep 2016
Posts: 4
Location: Saudi Arabia
Concentration: Finance, Entrepreneurship
WE: Corporate Finance (Consumer Electronics)
Re: In the addition shown above, A, B, C, and D represent [#permalink] | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637612961505,
"lm_q2_score": 0.8596637612961505,
"openwebmath_perplexity": 1315.5501724991552,
"openwebmath_score": 0.7710837125778198,
"tags": null,
"url": "https://gmatclub.com/forum/in-the-addition-shown-above-a-b-c-and-d-represent-161656.html"
} |
### Show Tags
26 May 2018, 03:22
mikemcgarry. Can you please explain Sarthak's query as i have the same query but unfortunately i didn't get from the reply you gave to him.
Intern
Joined: 11 May 2018
Posts: 5
Re: In the addition shown above, A, B, C, and D represent [#permalink]
### Show Tags
15 Jun 2018, 15:24
Do we not consider negative numbers in this problem at all?
Math Expert
Joined: 02 Sep 2009
Posts: 50727
Re: In the addition shown above, A, B, C, and D represent [#permalink]
### Show Tags
15 Jun 2018, 21:55
rohithtv89 wrote:
Do we not consider negative numbers in this problem at all?
No. We are given that A, B, C, and D represent the nonzero digits. There are 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
_________________
Re: In the addition shown above, A, B, C, and D represent &nbs [#permalink] 15 Jun 2018, 21:55
Display posts from previous: Sort by | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637612961505,
"lm_q2_score": 0.8596637612961505,
"openwebmath_perplexity": 1315.5501724991552,
"openwebmath_score": 0.7710837125778198,
"tags": null,
"url": "https://gmatclub.com/forum/in-the-addition-shown-above-a-b-c-and-d-represent-161656.html"
} |
GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 19 Aug 2018, 14:16
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# A photography dealer ordered 60 Model X cameras to be sold
Author Message
TAGS:
### Hide Tags
Intern
Joined: 26 Jul 2010
Posts: 24
A photography dealer ordered 60 Model X cameras to be sold [#permalink]
### Show Tags
06 May 2012, 10:53
21
129
00:00
Difficulty:
95% (hard)
Question Stats:
58% (03:01) correct 42% (03:26) wrong based on 2390 sessions
### HideShow timer Statistics | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637523076225,
"lm_q2_score": 0.8596637523076225,
"openwebmath_perplexity": 8908.812152197419,
"openwebmath_score": 0.5564415454864502,
"tags": null,
"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html"
} |
58% (03:01) correct 42% (03:26) wrong based on 2390 sessions
### HideShow timer Statistics
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit ##### Most Helpful Expert Reply Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8195 Location: Pune, India Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 24 Jun 2012, 22:20 61 38 pgmat wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras?
A. 7% loss
B. 13% loss
C. 7% profit
D. 13% profit
E. 15% profit
Use weighted avgs for a quick solution:
On 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1
Avg Profit/Loss = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit
For more on weighted avgs, check: http://www.veritasprep.com/blog/2011/03 ... -averages/
_________________
Karishma
Veritas Prep GMAT Instructor | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637523076225,
"lm_q2_score": 0.8596637523076225,
"openwebmath_perplexity": 8908.812152197419,
"openwebmath_score": 0.5564415454864502,
"tags": null,
"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html"
} |
Karishma
Veritas Prep GMAT Instructor
Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! ##### Most Helpful Community Reply Manager Status: mba here i come! Joined: 07 Aug 2011 Posts: 230 A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags Updated on: 25 Feb 2016, 08:28 98 27 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. suppose total cameras = 10 and each costs$10
total profit = $2*9 -$5*1 = $13 or 13% (with$100 cost)
_________________
press +1 Kudos to appreciate posts
Originally posted by MBAhereIcome on 23 Jun 2012, 07:31.
Last edited by MBAhereIcome on 25 Feb 2016, 08:28, edited 1 time in total.
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 48037
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
### Show Tags
06 May 2012, 11:07
25
1
24
pgmat wrote:
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Total cost 60*($250/1.2)=50*250;
# of cameras sold is 60-6=54 total revenue is 54*250;
# of cameras returned is 6 total refund 6*(250/1.2)*0.5;
So, total income 54*250+ 6*(250/1.2)*0.5
The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%
_________________
Manager
Joined: 02 Jun 2011
Posts: 132
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637523076225,
"lm_q2_score": 0.8596637523076225,
"openwebmath_perplexity": 8908.812152197419,
"openwebmath_score": 0.5564415454864502,
"tags": null,
"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html"
} |
### Show Tags
22 Jun 2012, 14:14
Bunuel wrote:
pgmat wrote:
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Total cost$60*250/1.2=50*250;
# of cameras sold is 60-6=54 total revenue is 54*250;
# of cameras returned is 6 total refund 6*(250/1.2)*0.5;
So, total income 54*250+ 6*(250/1.2)*0.5
The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%
Dear Bunuel,
Ths is a doubt i face with %ages, in ths quest too.
could you pls tell where i m going wrong?
when 20% mark up for initial cost is given , how to calculate it?
250 - 20/100*250 or to take if 120 is 250 then how much is 100?
how have you arrived at 1.2?
Math Expert
Joined: 02 Sep 2009
Posts: 48037
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
### Show Tags
23 Jun 2012, 04:06
3
2
kashishh wrote:
Bunuel wrote:
pgmat wrote:
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Total cost 60*$250/1.2=50*250;
# of cameras sold is 60-6=54 total revenue is 54*250;
# of cameras returned is 6 total refund 6*(250/1.2)*0.5;
So, total income 54*250+ 6*(250/1.2)*0.5 | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637523076225,
"lm_q2_score": 0.8596637523076225,
"openwebmath_perplexity": 8908.812152197419,
"openwebmath_score": 0.5564415454864502,
"tags": null,
"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html"
} |
The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%
Dear Bunuel,
Ths is a doubt i face with %ages, in ths quest too.
could you pls tell where i m going wrong?
when 20% mark up for initial cost is given , how to calculate it?
250 - 20/100*250 or to take if 120 is 250 then how much is 100?
how have you arrived at 1.2? | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637523076225,
"lm_q2_score": 0.8596637523076225,
"openwebmath_perplexity": 8908.812152197419,
"openwebmath_score": 0.5564415454864502,
"tags": null,
"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html"
} |
If it's given that the selling price is $250 and the markup over the initial cost is 20%, then: {Cost}+0.2{Cost}=1.2*{Cost}={Selling price} --> 1.2*{Cost}=$250 --> {Cost}=$250/1.2. Hope it's clear. _________________ Intern Joined: 28 Feb 2012 Posts: 21 GMAT 1: 700 Q48 V39 WE: Information Technology (Computer Software) Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 24 Jun 2012, 02:06 1 MBAhereIcome wrote: 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. suppose total cameras = 10, out of which 9 gave 20% profit suppose 100 was the cost of each. total profit 100*10=1000 profit from 9 = 20*9 = 180 loss from 1 = 50*1 = 50 total profit = 180-50 = 130 so, 130 profit from a cost of 1000. that is 13% profit. Your method seems easy, but I kind of got lost with this - 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. Could you please explain, how you deduced this? Thanks. Manager Joined: 02 Jun 2011 Posts: 132 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 24 Jun 2012, 12:26 1 2 gmatDeep wrote: MBAhereIcome wrote: 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. suppose total cameras = 10, out of which 9 gave 20% profit suppose 100 was the cost of each. total profit 100*10=1000 profit from 9 = 20*9 = 180 loss from 1 = 50*1 = 50 total profit = 180-50 = 130 so, 130 profit from a cost of 1000. that is 13% profit. Your method seems easy, but I kind of got lost with this - 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. Could you please explain, how you deduced this? Thanks. yess.. 54:6 = 9:1 mark up is of 20% = profit earned will be 20% VP Joined: 02 Jul 2012 Posts: 1192 Location: India Concentration: Strategy GMAT 1: 740 Q49 V42 GPA: 3.8 WE: Engineering (Energy and Utilities) Re: Of the cameras ordered, 6 were never sold and were returned [#permalink] ### Show | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637523076225,
"lm_q2_score": 0.8596637523076225,
"openwebmath_perplexity": 8908.812152197419,
"openwebmath_score": 0.5564415454864502,
"tags": null,
"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html"
} |
and Utilities) Re: Of the cameras ordered, 6 were never sold and were returned [#permalink] ### Show Tags 16 Nov 2012, 23:15 15 4 carcass wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, whichrepresents a 20 percent markup over the dealer's initialcost for each camera. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637523076225,
"lm_q2_score": 0.8596637523076225,
"openwebmath_perplexity": 8908.812152197419,
"openwebmath_score": 0.5564415454864502,
"tags": null,
"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html"
} |
Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer's initial cost for the 60 cameras? | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637523076225,
"lm_q2_score": 0.8596637523076225,
"openwebmath_perplexity": 8908.812152197419,
"openwebmath_score": 0.5564415454864502,
"tags": null,
"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html"
} |
(A) 7% loss
(B) 13% loss
(C) 7% profit
(D) 13% profit
(E) 15% profit
have fun. is not difficult to conceptualize but is difficult because if you are not organized and clear you are get in trouble.
I would not show the OA but the rules are stringent. either way try on your own
Cost to dealer for 60 cameras = $$\frac{250}{1.2}*60$$ = $12,500 Revenue for 54 cameras = 54*250 =$13,500
Revenue from 6 cameras = $$\frac{250}{2.4}*6$$ = $625 Total Revenue =$14,125
Profit percent = $$\frac{14125-12500}{12500}$$ = 13%
Otherwise, If one were short on time and I had to guess, One can eliminate A,B & C by looking at the problem itself and improve my chances of guessing the right answer to 50% between D & E.
Kudos Please... If my post helped.
_________________
Did you find this post helpful?... Please let me know through the Kudos button.
Thanks To The Almighty - My GMAT Debrief
GMAT Reading Comprehension: 7 Most Common Passage Types
VP
Joined: 24 Jul 2011
Posts: 1459
GMAT 1: 780 Q51 V48
GRE 1: Q800 V740
Re: Of the cameras ordered, 6 were never sold and were returned [#permalink]
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637523076225,
"lm_q2_score": 0.8596637523076225,
"openwebmath_perplexity": 8908.812152197419,
"openwebmath_score": 0.5564415454864502,
"tags": null,
"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html"
} |
### Show Tags
16 Nov 2012, 23:27
1
60 cameras were ordered to be sold at a 20% markup for $250 each => Dealer's cost for each camera =$208
=> Cost of all 60 cameras = $60*208 6 cameras were not sold and returned for a refund of 50% of the dealer's cost => The amount the dealer got in returns = 50% of 6*208 =$624
The rest of the cameras (60-6 = 54 in number) were sold
=> Revenue made from selling the cameras = $54 * 250 For the 60 cameras, dealer's profit = 100* [54*(250-208) + (624-1248)]/[(60*200)] =100* (14100-12000)/12000 = ~13% (the error arises because 250/1.2 is not exactly 208 but 208.33) Option (D) _________________ GyanOne | Top MBA Rankings and MBA Admissions Blog Top MBA Admissions Consulting | Top MiM Admissions Consulting Premium MBA Essay Review|Best MBA Interview Preparation|Exclusive GMAT coaching Get a FREE Detailed MBA Profile Evaluation | Call us now +91 98998 31738 Director Status: Done with formalities.. and back.. Joined: 15 Sep 2012 Posts: 616 Location: India Concentration: Strategy, General Management Schools: Olin - Wash U - Class of 2015 WE: Information Technology (Computer Software) Re: Of the cameras ordered, 6 were never sold and were returned [#permalink] ### Show Tags 16 Nov 2012, 23:48 24 13 carcass wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, whichrepresents a 20 percent markup over the dealer's initialcost for each camera.
Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer's initial cost for the 60 cameras?
(A) 7% loss
(B) 13% loss
(C) 7% profit
(D) 13% profit
(E) 15% profit
have fun. is not difficult to conceptualize but is difficult because if you are not organized and clear you are get in trouble.
I would not show the OA but the rules are stringent. either way try on your own | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637523076225,
"lm_q2_score": 0.8596637523076225,
"openwebmath_perplexity": 8908.812152197419,
"openwebmath_score": 0.5564415454864502,
"tags": null,
"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html"
} |
actually, if one could visualize this problem properly, it could be solved in less than 30 secs without getting into any dirty calculations.
Here is my 30 sec approach:
Notice that it is actually a wighted ratio problem, where 9 parts earned 20% profit and 1 part earned 50% loss. (54 for profit, 6 for loss).
Hence overall profit/loss = $$(9*20 -1*50) /(9+1) = 13$$
Ans D it is!
_________________
Lets Kudos!!!
Black Friday Debrief
Intern
Joined: 16 Nov 2009
Posts: 7
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
### Show Tags
24 Nov 2012, 11:16
2
Hi,
For those using normal way to solve this problem, I think that we can avoid huge calculation by just replace Cost by letter C and put it in the final formula:
Income: 54*1.2C + 6*0.5C
Cost: 60*C
Then percent =(Income-cost)/cost = (54*1.2C + 6*0.5C - 60*C)/60*C. We can see that we can eliminate C from the fraction --> percent = (54*1.2 + 6*0.5 - 60)/60 = (54*1.2 + 6*0.5)/60 - 1. Here, we can see clearly the weighted average part of the problem or we can simply calculate the fraction to reach the result.
Intern
Status: K... M. G...
Joined: 22 Oct 2012
Posts: 36
GMAT Date: 08-27-2013
GPA: 3.8
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
### Show Tags
27 May 2013, 12:12
Bunuel wrote:
pgmat wrote:
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Total cost 60*($250/1.2)=50*250;
# of cameras sold is 60-6=54 total revenue is 54*250;
# of cameras returned is 6 total refund 6*(250/1.2)*0.5;
So, total income 54*250+ 6*(250/1.2)*0.5 | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637523076225,
"lm_q2_score": 0.8596637523076225,
"openwebmath_perplexity": 8908.812152197419,
"openwebmath_score": 0.5564415454864502,
"tags": null,
"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html"
} |
The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%
this must be kind of awkward question but i got no option other than asking you.
20% markup over the dealer’s initial cost for each camera- I made a wrong assumption , but reducing 20%from 250 & got 200 as an initial cost for each. please explain that "Total cost 60*($250/1.2)=50*250;" Math Expert Joined: 02 Sep 2009 Posts: 48037 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 27 May 2013, 12:19 3 1 FTGNGU wrote: Bunuel wrote: pgmat wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras?
A. 7% loss
B. 13% loss
C. 7% profit
D. 13% profit
E. 15% profit
Total cost 60*($250/1.2)=50*250; # of cameras sold is 60-6=54 total revenue is 54*250; # of cameras returned is 6 total refund 6*(250/1.2)*0.5; So, total income 54*250+ 6*(250/1.2)*0.5 The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13% Answer: D. this must be kind of awkward question but i got no option other than asking you. 20% markup over the dealer’s initial cost for each camera- I made a wrong assumption , but reducing 20%from 250 & got 200 as an initial cost for each. please explain that "Total cost 60*($250/1.2)=50*250;"
(Cost per unit) + 0.2*(Cost per unit) = $250 1.2*(Cost per unit) =$250
(Cost per unit) = $250/1.2 Total cost for 60 units = 60*(Cost per unit) = 60*($250/1.2) = 50*250.
Hope it's clear.
_________________
Intern
Joined: 02 Mar 2010
Posts: 19
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637523076225,
"lm_q2_score": 0.8596637523076225,
"openwebmath_perplexity": 8908.812152197419,
"openwebmath_score": 0.5564415454864502,
"tags": null,
"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html"
} |
07 Apr 2014, 19:23
4 | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637523076225,
"lm_q2_score": 0.8596637523076225,
"openwebmath_perplexity": 8908.812152197419,
"openwebmath_score": 0.5564415454864502,
"tags": null,
"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html"
} |
Because the answers are in percentages, I thought not to worry about the $amounts and just focus on the relationships: Let C be the total cost of all 60 cameras. Originally, the dealer thought to sell all these 60 cameras at 20% profit or for (1.2)C. However, he sold only (60-6)=54 or 90% of cameras at this price. So revenue from these cameras = (0.9)(1.2)C = (1.08)C For the remaining 10%, he got a refund of 50% of cost or (0.5)C. So total refund = (0.1)(0.5)C = (0.05)C Therefore, total revenue in terms of original cost = (1.08 + 0.05)C = 1.13C or 13% profit. So D is the correct ans. Intern Status: 1st attempt: Can I do it? Joined: 09 Jun 2013 Posts: 1 Location: India Concentration: Healthcare, Marketing GMAT Date: 05-30-2014 GPA: 2.99 WE: Pharmaceuticals (Pharmaceuticals and Biotech) Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 27 May 2014, 10:33 2 1 {(54*20) + (6*(-50))}/ 60 This gives 13% profit as answer. 20% proft on 54 cameras and 50% loss on 6 cameras. Use weighted averages here. Rest of the information in the question is misguiding. Intern Joined: 09 Feb 2013 Posts: 26 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 29 Jun 2014, 23:52 My soln is : Dealer's per unit initial cost =250 -20/100*250 = 200 Total initial cost = 60*200=12000 Total profit = (54*250+6*100) - 12000 = 2100 Profit % as a percent of dealer's initial cost = 2100/12000*100 = 17.5 % Can you please tell me where I went wrong ? Posted from my mobile device Math Expert Joined: 02 Sep 2009 Posts: 48037 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 30 Jun 2014, 02:50 kshitij89 wrote: My soln is : Dealer's per unit initial cost =250 -20/100*250 = 200 Total initial cost = 60*200=12000 Total profit = (54*250+6*100) - 12000 = 2100 Profit % as a percent of dealer's initial cost = 2100/12000*100 = 17.5 % Can you please tell me where I went wrong ? Posted from my mobile | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637523076225,
"lm_q2_score": 0.8596637523076225,
"openwebmath_perplexity": 8908.812152197419,
"openwebmath_score": 0.5564415454864502,
"tags": null,
"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html"
} |
cost = 2100/12000*100 = 17.5 % Can you please tell me where I went wrong ? Posted from my mobile device The cost per unit is not 0.8*250 = 200, it's 250/1.2 = ~208. Markup is calculated on the cost value (check here: http://gmatclub.com/forum/a-photography ... l#p1229596). Hope it helps. _________________ Manager Joined: 28 Dec 2013 Posts: 68 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 21 Aug 2014, 08:48 VeritasPrepKarishma wrote: pgmat wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637523076225,
"lm_q2_score": 0.8596637523076225,
"openwebmath_perplexity": 8908.812152197419,
"openwebmath_score": 0.5564415454864502,
"tags": null,
"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html"
} |
A. 7% loss
B. 13% loss
C. 7% profit
D. 13% profit
E. 15% profit
Use weighted avgs for a quick solution:
On 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1
Avg Profit/Loss = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit
For more on weighted avgs, check: http://www.veritasprep.com/blog/2011/03 ... -averages/
why are we dividing by 10?
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8195
Location: Pune, India
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637523076225,
"lm_q2_score": 0.8596637523076225,
"openwebmath_perplexity": 8908.812152197419,
"openwebmath_score": 0.5564415454864502,
"tags": null,
"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html"
} |
### Show Tags
24 Aug 2014, 21:54
1
sagnik242 wrote:
VeritasPrepKarishma wrote:
pgmat wrote:
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Use weighted avgs for a quick solution: On 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1 Avg Profit/Loss = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit For more on weighted avgs, check: http://www.veritasprep.com/blog/2011/03 ... -averages/ why are we dividing by 10? Weighted Average Formula: Cavg = (C1*w1 + C2*w2)/(w1 + w2) C1 and C2 represent the quantity which we want to average so they will be profit/loss here. C1 = 20% = .2 C2 = -50% (loss) = -.5 The weights, given by w1 and w2, are the cost prices. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1 So w1 =9 and w2 = 1 Avg Profit/Loss = (.2*9 + (-.5)*1)/(9 + 1) = (.2*9 + (-.5)*1)/10 Get more details on this concept from the link given in my post above. _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >
GMAT self-study has never been more personalized or more fun. Try ORION Free!
Re: A photography dealer ordered 60 Model X cameras to be sold &nbs [#permalink] 24 Aug 2014, 21:54
Go to page 1 2 3 Next [ 57 posts ]
Display posts from previous: Sort by
# Events & Promotions | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637523076225,
"lm_q2_score": 0.8596637523076225,
"openwebmath_perplexity": 8908.812152197419,
"openwebmath_score": 0.5564415454864502,
"tags": null,
"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html"
} |
Display posts from previous: Sort by
# Events & Promotions
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637523076225,
"lm_q2_score": 0.8596637523076225,
"openwebmath_perplexity": 8908.812152197419,
"openwebmath_score": 0.5564415454864502,
"tags": null,
"url": "https://gmatclub.com/forum/a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html"
} |
It is currently 20 Nov 2017, 23:39
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# If x is an integer then x(x-1)(x-k) must be evenly divisible
Author Message
TAGS:
### Hide Tags
Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 532
Kudos [?]: 4216 [1], given: 217
Location: United Kingdom
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)
If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
### Show Tags
31 Jan 2012, 15:49
1
KUDOS
50
This post was
BOOKMARKED
00:00
Difficulty:
55% (hard)
Question Stats:
64% (01:19) correct 36% (01:25) wrong based on 1112 sessions
### HideShow timer Statistics
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A. -4
B. -2
C. -1
D. 2
E. 5
[Reveal] Spoiler:
The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?
Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.
[Reveal] Spoiler: OA
_________________
Best Regards,
E.
MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730
Kudos [?]: 4216 [1], given: 217
Math Expert
Joined: 02 Sep 2009
Posts: 42275
Kudos [?]: 132868 [13], given: 12389
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637451167997,
"lm_q2_score": 0.8596637451167997,
"openwebmath_perplexity": 2078.5322086005,
"openwebmath_score": 0.5184522271156311,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html?fl=similar"
} |
Math Expert
Joined: 02 Sep 2009
Posts: 42275
Kudos [?]: 132868 [13], given: 12389
### Show Tags
31 Jan 2012, 16:06
13
KUDOS
Expert's post
31
This post was
BOOKMARKED
enigma123 wrote:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A)-4
B)-2
C)-1
D) 2
E) 5
The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?
Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.
We have the product of 3 integers: (x-1)x(x-k).
Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.
Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.
30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.
Hope it helps.
_________________
Kudos [?]: 132868 [13], given: 12389
Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 532
Kudos [?]: 4216 [0], given: 217
Location: United Kingdom
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)
Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637451167997,
"lm_q2_score": 0.8596637451167997,
"openwebmath_perplexity": 2078.5322086005,
"openwebmath_score": 0.5184522271156311,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html?fl=similar"
} |
### Show Tags
31 Jan 2012, 16:09
Thanks very much for a thorough explanation.
_________________
Best Regards,
E.
MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730
Kudos [?]: 4216 [0], given: 217
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4491
Kudos [?]: 8757 [15], given: 105
### Show Tags
31 Jan 2012, 16:17
15
KUDOS
Expert's post
10
This post was
BOOKMARKED
Hi, there. I'm happy to help with this.
The rule that the product of three consecutive integers is a good start, but not the be all and end all.
number = x(x – 1)(x – k)
So far, we have integer x and one less that it (x - 1), so we could go down one more, or up one from x --
k = 2 ----> x(x – 1)(x – 2)
k = -1 ----> x(x – 1)(x + 1)
Now, we don't know which of the three factors are divisible by 3 -- x, or (x - 1), or the (x - k). If it's either of the first two, then we're golden, and k doesn't matter. But pretend that neither x nor (x - 1) is divisible by 3, then we are dependent on that last factor. Well, if (x - 2) is a multiple of three, we should be able to add or subtract three and still get a multiple of three.
(x - 2) - 3 = (x - 5)
(x - 5) - 3 = (x - 8)
(x - 2) + 3 = (x + 1), which we have already
(x + 1) + 3 = (x + 4)
(x + 3) + 3 = (x + 7)
So, for divisibility purposes, (x - 8), (x - 5), (x - 2), (x + 1), (x + 4), (x + 7) are all equivalent -- if any one of them is a multiple of three, all the others are. (You can check that the difference between any two is a multiple of 3.) BTW, if they are not divisible by three, then they all would have equal remainders if divided by three.
Back to the question:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A)-4
B)-2
C)-1
D) 2
E) 5
All of those choices give us a term on our list except for (B) -2.
BTW, notice all the answer choices are spaced apart by three except for (B).
Does that make sense? Please do not hesitate to ask if you have any questions. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637451167997,
"lm_q2_score": 0.8596637451167997,
"openwebmath_perplexity": 2078.5322086005,
"openwebmath_score": 0.5184522271156311,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html?fl=similar"
} |
Does that make sense? Please do not hesitate to ask if you have any questions.
Mike
_________________
Mike McGarry
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Kudos [?]: 8757 [15], given: 105
Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 532
Kudos [?]: 4216 [0], given: 217
Location: United Kingdom
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)
Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
### Show Tags
31 Jan 2012, 16:18
Thanks Mike. Really appreciate your solution too.
_________________
Best Regards,
E.
MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730
Kudos [?]: 4216 [0], given: 217
Intern
Joined: 31 Jan 2012
Posts: 1
Kudos [?]: 4 [4], given: 0
Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
### Show Tags
31 Jan 2012, 16:34
4
KUDOS
This is my 1st post finally thought of jumping in instead of just being an observer
I attacked this problem in a simple way. As it states it is divisible by 3
that means both x & (x-1) cannot be a multiple of 3 otherwise whatever the value of k it will be still divisible by 3
so plugging in number i chose 5 in this case you can establish answer is -2 does not fit...
Kudos [?]: 4 [4], given: 0
Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 532
Kudos [?]: 4216 [0], given: 217
Location: United Kingdom
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)
Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
### Show Tags
31 Jan 2012, 16:37
Thanks and welcome to GMAT Club Azim. I am sure you will have a great experience from a very helpful community.
_________________
Best Regards,
E.
MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730
Kudos [?]: 4216 [0], given: 217
Intern
Joined: 31 May 2012
Posts: 11 | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637451167997,
"lm_q2_score": 0.8596637451167997,
"openwebmath_perplexity": 2078.5322086005,
"openwebmath_score": 0.5184522271156311,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html?fl=similar"
} |
Kudos [?]: 4216 [0], given: 217
Intern
Joined: 31 May 2012
Posts: 11
Kudos [?]: 14 [0], given: 8
### Show Tags
04 Oct 2012, 18:59
Bunuel wrote:
enigma123 wrote:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A)-4
B)-2
C)-1
D) 2
E) 5
The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?
Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.
We have the product of 3 integers: (x-1)x(x-k).
Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.
Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.
30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.
Hope it helps.
Bunuel,
Would this approach work for all integer divisors (NOT the 30 sec approach)?
Say the divisor is 4 and the choices had four terms instead of three, e.g. (x-1)(x-2)(x+k)(x+1)?
Thanks,
Kudos [?]: 14 [0], given: 8
Director
Joined: 22 Mar 2011
Posts: 610
Kudos [?]: 1074 [8], given: 43
WE: Science (Education)
Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
### Show Tags | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637451167997,
"lm_q2_score": 0.8596637451167997,
"openwebmath_perplexity": 2078.5322086005,
"openwebmath_score": 0.5184522271156311,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html?fl=similar"
} |
### Show Tags
04 Oct 2012, 23:10
8
KUDOS
2
This post was
BOOKMARKED
enigma123 wrote:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A. -4
B. -2
C. -1
D. 2
E. 5
The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?
Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.
Since this is a multiple choice GMAT question, you can pick a particular value for x such that neither x, nor x-1 is divisible by 3 and start checking the answers.
In the given situation, choose for example x = 2 and check when 2 - k is not divisible by 3.
(A) 2 - (-4) = 6 NO
(B) 2 - (-2) = 4 BINGO!
_________________
PhD in Applied Mathematics
Love GMAT Quant questions and running.
Kudos [?]: 1074 [8], given: 43
Intern
Joined: 10 Jul 2012
Posts: 14
Kudos [?]: 7 [0], given: 0
GMAT Date: 10-20-2012
Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
### Show Tags
05 Oct 2012, 03:37
I too solved it using a value for x =2, but I am not sure if is it better to solve using value for such questions or otherwise.
Kudos [?]: 7 [0], given: 0
Director
Joined: 22 Mar 2011
Posts: 610
Kudos [?]: 1074 [1], given: 43
WE: Science (Education)
Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
### Show Tags
05 Oct 2012, 03:56
1
KUDOS
Avantika5 wrote:
I too solved it using a value for x =2, but I am not sure if is it better to solve using value for such questions or otherwise. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637451167997,
"lm_q2_score": 0.8596637451167997,
"openwebmath_perplexity": 2078.5322086005,
"openwebmath_score": 0.5184522271156311,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html?fl=similar"
} |
On the GMAT, is definitely the fastest way to solve it. Being a multiple choice question, you can be sure that there is a unique correct answer. And in the given situation, the only issue is to choose for x values such that neither x, nor x - 1 is divisible by 3.
It won't harm to understand and know to use the properties of consecutive integers presented in the other posts . They can be useful any time.
_________________
PhD in Applied Mathematics
Love GMAT Quant questions and running.
Kudos [?]: 1074 [1], given: 43
Intern
Joined: 10 Jul 2012
Posts: 14
Kudos [?]: 7 [0], given: 0
GMAT Date: 10-20-2012
Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
### Show Tags
06 Oct 2012, 01:10
Thanks EvaJager, I was always thought it is not good way to solve and I should learn the better way.
Thanks a lot...
Kudos [?]: 7 [0], given: 0
Director
Joined: 22 Mar 2011
Posts: 610
Kudos [?]: 1074 [0], given: 43
WE: Science (Education)
Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
### Show Tags
06 Oct 2012, 02:10
Avantika5 wrote:
Thanks EvaJager, I was always thought it is not good way to solve and I should learn the better way.
Thanks a lot...
Better is a relative word...Mathematicians always try to prove and justify everything in a formal, logical way.
But GMAT is not testing mathematical abilities per se. If they wanted so, the questions would have been open and not multiple choice.
Have a flexible mind, think out of the box. GMAT is not a contest for the most beautiful, elegant, mathematical solution...
Get the correct answer as quickly as possible, and go to the next question without any feeling of guilt...:O)
Though, as I said, try to understand the other properties of the integer numbers, they can become handy and also, because they are so beautiful! Isn't Mathematics wonderful?
_________________
PhD in Applied Mathematics
Love GMAT Quant questions and running. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637451167997,
"lm_q2_score": 0.8596637451167997,
"openwebmath_perplexity": 2078.5322086005,
"openwebmath_score": 0.5184522271156311,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html?fl=similar"
} |
PhD in Applied Mathematics
Love GMAT Quant questions and running.
Last edited by EvaJager on 06 Oct 2012, 02:46, edited 1 time in total.
Kudos [?]: 1074 [0], given: 43
Manager
Status: Fighting hard
Joined: 04 Jul 2011
Posts: 69
Kudos [?]: 83 [0], given: 87
GMAT Date: 10-01-2012
Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
### Show Tags
06 Oct 2012, 02:25
1. k=-4, x+4=y so x=y-4. Now ----- (y-4)(y-5)(y). Gives number divisible by 3 in all cases.
2. k=-2, , x+2=y so x=y-2. Now ----- (y-2)(y-3)(y). Not applicable when y is 5 or when x is 7
3. k=-1, , x+1=y so x=y-1. Now ----- (y-1)(y-2)(y). Consecutive 3 integers. Divisible by 3
4. k=2, , x-2=y so x=y+2. Now ----- (y+2)(y+1)(y). Consecutive 3 integers. Divisible by 3
5. k=5, , x-5=y so x=y+5. Now ----- (y+5)(y+4)(y). Consecutive 3 integers. Divisible by 3
_________________
I will rather do nothing than be busy doing nothing - Zen saying
Kudos [?]: 83 [0], given: 87
Senior Manager
Joined: 15 Sep 2011
Posts: 358
Kudos [?]: 416 [0], given: 45
Location: United States
WE: Corporate Finance (Manufacturing)
### Show Tags
16 Feb 2013, 12:34
mikemcgarry wrote:
Hi, there. I'm happy to help with this.
The rule that the product of three consecutive integers is a good start, but not the be all and end all.
number = x(x – 1)(x – k)
So far, we have integer x and one less that it (x - 1), so we could go down one more, or up one from x --
k = 2 ----> x(x – 1)(x – 2)
k = -1 ----> x(x – 1)(x + 1)
Now, we don't know which of the three factors are divisible by 3 -- x, or (x - 1), or the (x - k). If it's either of the first two, then we're golden, and k doesn't matter. But pretend that neither x nor (x - 1) is divisible by 3, then we are dependent on that last factor. Well, if (x - 2) is a multiple of three, we should be able to add or subtract three and still get a multiple of three.
(x - 2) - 3 = (x - 5)
(x - 5) - 3 = (x - 8)
(x - 2) + 3 = (x + 1), which we have already | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637451167997,
"lm_q2_score": 0.8596637451167997,
"openwebmath_perplexity": 2078.5322086005,
"openwebmath_score": 0.5184522271156311,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html?fl=similar"
} |
(x - 2) - 3 = (x - 5)
(x - 5) - 3 = (x - 8)
(x - 2) + 3 = (x + 1), which we have already
(x + 1) + 3 = (x + 4)
(x + 3) + 3 = (x + 7)
So, for divisibility purposes, (x - 8), (x - 5), (x - 2), (x + 1), (x + 4), (x + 7) are all equivalent -- if any one of them is a multiple of three, all the others are. (You can check that the difference between any two is a multiple of 3.) BTW, if they are not divisible by three, then they all would have equal remainders if divided by three.
Back to the question:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A)-4
B)-2
C)-1
D) 2
E) 5
All of those choices give us a term on our list except for (B) -2.
BTW, notice all the answer choices are spaced apart by three except for (B).
Does that make sense? Please do not hesitate to ask if you have any questions.
Mike
Great posts everyone, including you Mike, who I am now quoting, but I would like to add that..
The first two integers for each answer choice cannot be divisible by 3 because all the answers choices would then be correct, thereby making the question invalid. Therefore, the last integer (x+k) must be divisible by 3, which helps in the theoretical approach...
It appears that was left out
Kudos [?]: 416 [0], given: 45
Current Student
Joined: 06 Sep 2013
Posts: 1972
Kudos [?]: 742 [0], given: 355
Concentration: Finance
Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
### Show Tags
17 Oct 2013, 15:25
enigma123 wrote:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A. -4
B. -2
C. -1
D. 2
E. 5
The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?
Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637451167997,
"lm_q2_score": 0.8596637451167997,
"openwebmath_perplexity": 2078.5322086005,
"openwebmath_score": 0.5184522271156311,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html?fl=similar"
} |
Hi enigma123. Thank you for the nice question. But you might wanna put a spoiler on this statement The OA is B so that others can solve the question in real conditions. Thanks
Kudos [?]: 742 [0], given: 355
Manager
Joined: 27 Aug 2014
Posts: 102
Kudos [?]: 119 [10], given: 19
Concentration: Finance, Strategy
GPA: 3.9
WE: Analyst (Energy and Utilities)
Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
### Show Tags
27 Oct 2014, 06:45
10
KUDOS
2
This post was
BOOKMARKED
I tried the problem in a different way:
for any number to be divisible by 3, the sum of the integers in the number should be a factor of 3.
Taking the sum of x, (x-1) and (x-k) we have:
x+x-1+x-k = 3x-1-k
now looking at the choices
k = -4 => sum = 3x+3 --> divisible by 3
k = -2 => sum = 3x+1 --> not divisible by 3
k = -1 => sum = 3x --> divisible by 3
k = 2 => sum = 3x-3 --> divisible by 3
k = 5 ==> sum = 3x-6 --> divisible by 3
Kudos [?]: 119 [10], given: 19
Intern
Joined: 29 Apr 2015
Posts: 2
Kudos [?]: 5 [0], given: 47
Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
### Show Tags
17 Jun 2015, 02:13
Hi there,
The answer is right, however my explanation is quite differently:
1. x(x-1)(x-k)
2. (x^2-1) (x-k)
3. x^3-x^2k-x^2+xk
4. x-xk
Plugg-in x=1
a. 1-1(-4) = 1-(-4) = 5
b. 1-1(-2) = 1-(-2) = 3
c. 1-1(-1) = 1-(-1) = 2
d. 1-1(2) = 1-(2) = -1
e. 1-1(5) = 1-(5) = -4
Am I correct in my explanation?
Kudos [?]: 5 [0], given: 47
SVP
Joined: 08 Jul 2010
Posts: 1851
Kudos [?]: 2345 [3], given: 51
Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
### Show Tags
17 Jun 2015, 05:49
3
KUDOS
Expert's post
santorasantu wrote:
Quote:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A. -4
B. -2
C. -1
D. 2
E. 5
I tried the problem in a different way: | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637451167997,
"lm_q2_score": 0.8596637451167997,
"openwebmath_perplexity": 2078.5322086005,
"openwebmath_score": 0.5184522271156311,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html?fl=similar"
} |
A. -4
B. -2
C. -1
D. 2
E. 5
I tried the problem in a different way:
for any number to be divisible by 3, the sum of the integers in the number should be a factor of 3.
Taking the sum of x, (x-1) and (x-k) we have:
x+x-1+x-k = 3x-1-k
now looking at the choices
k = -4 => sum = 3x+3 --> divisible by 3
k = -2 => sum = 3x+1 --> not divisible by 3
k = -1 => sum = 3x --> divisible by 3
k = 2 => sum = 3x-3 --> divisible by 3
k = 5 ==> sum = 3x-6 --> divisible by 3
Just refining the highlighted language and presenting another view to del with this problem
CONCEPT1:For any number to be divisible by 3, the sum of the Digits of the Number should be a Multiple of 3.
CONCEPT2: Product of any three consecutive Integers always include one muliple of 3 hence product of any three consecutive Integers is always a Multiple of 3
CONCEPT3: If a Number "w" is a Multiple of 3 then any number at a difference of 3 or multiple of 3 from "w" will also be a multiple of 3 i.e. If w is a multiple of 3 then (w+3), (w-3), (w+6), (w-6) etc. will all be Multiples of 3
Here, I see that x(x – 1) is a product of two consecutive Integers but if another Number next to them is obtained then x(x – 1)(x – k) will certainly be a multiple of 3 [as per Concept2 mentioned above]
for x(x – 1)(x – k) to be a product of 3 consecutive integers,
(x – k) should be either (x - 2) i.e. k=2
OR
(x – k) should be either (x - 5) i.e. k=5 [Using Concept3]
OR
(x – k) should be either (x + 1) i.e. k=-1
OR
(x – k) should be either (x + 4) i.e. k=-4 [Using Concept3]
This Eliminates options A, C D and E
[Reveal] Spoiler:
B
_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html
22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
Kudos [?]: 2345 [3], given: 51 | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637451167997,
"lm_q2_score": 0.8596637451167997,
"openwebmath_perplexity": 2078.5322086005,
"openwebmath_score": 0.5184522271156311,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html?fl=similar"
} |
Kudos [?]: 2345 [3], given: 51
Senior Manager
Joined: 13 Oct 2016
Posts: 367
Kudos [?]: 402 [0], given: 40
GPA: 3.98
If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
### Show Tags
04 Nov 2016, 04:23
1
This post was
BOOKMARKED
enigma123 wrote:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A. -4
B. -2
C. -1
D. 2
E. 5
[Reveal] Spoiler:
The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?
Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.
A little bit faster approach.
We know from the start that product of $$3$$ consecutive integers is divisible by $$3$$, because it’s indeed a multiple of $$3$$.
So we have $$x(x+1)(x+2)$$ is definitely divisible by 3 and k=2 is our anchor point.
Now we can use principles of modular arithmetic to generate multiples of $$3$$. We can either add or subtract $$3$$ from our anchor number. And we got following string of numbers:
… -4, -1, 2, 5, 7 …
As you can see in this progression only answer B (-2) is absent.
Kudos [?]: 402 [0], given: 40
If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink] 04 Nov 2016, 04:23
Go to page 1 2 Next [ 22 posts ]
Display posts from previous: Sort by | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.8596637451167997,
"lm_q2_score": 0.8596637451167997,
"openwebmath_perplexity": 2078.5322086005,
"openwebmath_score": 0.5184522271156311,
"tags": null,
"url": "https://gmatclub.com/forum/if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html?fl=similar"
} |
## Is 0/0 undefined or indeterminate?
Discussions concerned with knowledge of measurement, properties, and relations quantities, theoretical or applied.
### Is 0/0 undefined or indeterminate?
I'm confused, my book says that it's undefined, but I've been reading on the internet that it's indeterminate.
krum
Forum Neophyte
Posts: 6
Joined: 08 Oct 2014
### Re: Is 0/0 undefined or indeterminate?
krum » August 16th, 2015, 7:37 pm wrote:I'm confused, my book says that it's undefined, but I've been reading on the internet that it's indeterminate.
The way I learned it in school:
a/0 is 'undefined' when a≠0,
a/0 is 'indeterminate' when a=0.
Thus, unless nomenclature has changed since I was in school, the answer to your question is 'indeterminate'.
Darby
Active Member
Posts: 1188
Joined: 14 Feb 2015
Location: Long Island, New York (USA)
### Re: Is 0/0 undefined or indeterminate?
Calculus books will refer to "indeterminate forms", because they play a special role in certain infinite series.
$0^{0}$ is one of them. As is
$\frac{0}{0} ,$
$\infty ^{\infty },$
$\frac{\infty }{\infty },$ and
$\infty ^{0}$
However, 1/0 is not an indeterminate form, and for that reason it can't be used in certain theorems. If those "undefined" a/0 things appear in certain limits of infinite series, math students can justifiably say the the answer is "infinity", and be technically correct. Depending on things like the textbook or the professor, the right answer may also be saying that the series (or limit) "diverges."
hyksos
Active Member
Posts: 1294
Joined: 28 Nov 2014
### Re: Is 0/0 undefined or indeterminate?
krum, | {
"domain": "sciencechatforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.97241471777321,
"lm_q1q2_score": 0.8596528101449367,
"lm_q2_score": 0.8840392832736084,
"openwebmath_perplexity": 1630.4931841615064,
"openwebmath_score": 0.8781976103782654,
"tags": null,
"url": "http://sciencechatforum.com/viewtopic.php?f=19&t=29316&p=303488&sid=7c166a4316456a6b426d7668e4965885"
} |
Posts: 1294
Joined: 28 Nov 2014
### Re: Is 0/0 undefined or indeterminate?
krum,
What's $\frac{x}{0}$? Okay, let's ask that question mathematically by:
1. Call the definition $k$.
2. Write that $k=\frac{x}{0}$.
3. Solve for $x$:
• $k=\frac{x}{0}$;
• $0{\times}k=0{\times}{\frac{x}{0}}$
• $0{\times}k=x$
• $x=0$.
4. Consider $x=0$.
• No definition $k$ would lead to a contradiction.
• So all possible definitions $k$ may be consistent.
• So we cannot determine a particular definition that $k$ must be.
• Call this inability to determine "indeterminate".
5. Consider $x{\neq}0$.
• Any definition $k$ would lead to a contradiction.
• So there is no possible definition $k$.
• Call this inability to define "undefined".
Therefore $\frac{x}{0}$ is:
1. indeterminate when $x=0$.
2. undefined when $x{\neq}0$.
Historically some mathematicians have failed to rigorously consider branching cases when they had fractions. For example, if their proof contained $\frac{x}{y}$ at some intermediate step but was later removed (e.g. by multiplication by $y$), then it may be hard to tell, but the proof may be flawed through reliance on $0{\times}{\frac{x}{0}}=x$. This hidden division-by-zero fallacy was the first major math fallacy that I was warned about.
Natural ChemE
Forum Moderator
Posts: 2754
Joined: 28 Dec 2009
Lomax, CanadysPeak, Darby liked this post
### Re: Is 0/0 undefined or indeterminate?
krum » Sun Aug 16, 2015 6:37 pm wrote:I'm confused, my book says that it's undefined, but I've been reading on the internet that it's indeterminate.
I hope this isn't off topic, but undefined suggests it can be, but just hasn't yet been defined. Would not undefinable be more accurate, in the context of the OP?
Watson
Resident Member
Posts: 4605
Joined: 19 Apr 2009
Location: Earth, middle of the top half, but only briefly each 24 hours.
### Re: Is 0/0 undefined or indeterminate? | {
"domain": "sciencechatforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.97241471777321,
"lm_q1q2_score": 0.8596528101449367,
"lm_q2_score": 0.8840392832736084,
"openwebmath_perplexity": 1630.4931841615064,
"openwebmath_score": 0.8781976103782654,
"tags": null,
"url": "http://sciencechatforum.com/viewtopic.php?f=19&t=29316&p=303488&sid=7c166a4316456a6b426d7668e4965885"
} |
### Re: Is 0/0 undefined or indeterminate?
Watson » August 17th, 2015, 12:58 pm wrote:
krum » Sun Aug 16, 2015 6:37 pm wrote:I'm confused, my book says that it's undefined, but I've been reading on the internet that it's indeterminate.
I hope this isn't off topic, but undefined suggests it can be, but just hasn't yet been defined. Would not undefinable be more accurate, in the context of the OP?
No. Classical Mathematics is actually quite clear and unambiguous on this point. It's been well covered in the previous posts.
Darby
Active Member
Posts: 1188
Joined: 14 Feb 2015
Location: Long Island, New York (USA)
### Re: Is 0/0 undefined or indeterminate?
To put it more philosophically, x/x means a ratio of things - litterboxes to cats, doorknobs to doors, etc. Zero isn't anything, it isn't a real quantity, it is simply an absence, viz. nothing. Seen clearly, it has no place in a ratio. When you have no cats or litterboxes, then their ratio is indeterminate because there are no things to relate to each other. When you have ten cats and no litterboxes, then you have a ratio that is undefined - 10/0. You can't define a ratio because you only have one sort of quantity. And a yard full of cat poop. Does this sound right?
Braininvat
Posts: 6470
Joined: 21 Jan 2014
Location: Black Hills
### Re: Is 0/0 undefined or indeterminate?
So it is undefinable. I'm not sure how this is somewhat the same as indeterminable.
Watson
Resident Member
Posts: 4605
Joined: 19 Apr 2009
Location: Earth, middle of the top half, but only briefly each 24 hours.
Natural ChemE liked this post
### Re: Is 0/0 undefined or indeterminate?
Thay are not the same ... they are different terms each with their own specific definitions.
The inelegant but tried and true school methodology of learning may come in handy herr ... if understanding is not instantly forthcoming, rely on memorization until it does. Worked for me when I needed it.
Darby
Active Member | {
"domain": "sciencechatforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.97241471777321,
"lm_q1q2_score": 0.8596528101449367,
"lm_q2_score": 0.8840392832736084,
"openwebmath_perplexity": 1630.4931841615064,
"openwebmath_score": 0.8781976103782654,
"tags": null,
"url": "http://sciencechatforum.com/viewtopic.php?f=19&t=29316&p=303488&sid=7c166a4316456a6b426d7668e4965885"
} |
Darby
Active Member
Posts: 1188
Joined: 14 Feb 2015
Location: Long Island, New York (USA)
### Re: Is 0/0 undefined or indeterminate?
Natural ChemE » Mon Aug 17, 2015 1:26 am wrote:krum,
What's $\frac{x}{0}$? Okay, let's ask that question mathematically by:
1. Call the definition $k$.
2. Write that $k=\frac{x}{0}$.
3. Solve for $x$:
• $k=\frac{x}{0}$;
• $0{\times}k=0{\times}{\frac{x}{0}}$
• $0{\times}k=x$
• $x=0$.
4. Consider $x=0$.
• No definition $k$ would lead to a contradiction.
• So all possible definitions $k$ may be consistent.
• So we cannot determine a particular definition that $k$ must be.
• Call this inability to determine "indeterminate".
5. Consider $x{\neq}0$.
• Any definition $k$ would lead to a contradiction.
• So there is no possible definition $k$.
• Call this inability to define "undefined".
Therefore $\frac{x}{0}$ is:
1. indeterminate when $x=0$.
2. undefined when $x{\neq}0$.
Historically some mathematicians have failed to rigorously consider branching cases when they had fractions. For example, if their proof contained $\frac{x}{y}$ at some intermediate step but was later removed (e.g. by multiplication by $y$), then it may be hard to tell, but the proof may be flawed through reliance on $0{\times}{\frac{x}{0}}=x$. This hidden division-by-zero fallacy was the first major math fallacy that I was warned about.
Outstanding! Perhaps the clearest explanation of this that I have seen.
Resident Expert
Posts: 5931
Joined: 31 Dec 2008
### Re: Is 0/0 undefined or indeterminate?
Watson » August 17th, 2015, 4:34 pm wrote:So it is undefinable. I'm not sure how this is somewhat the same as indeterminable.
Yup, "undefinable" would be a good way to think about it.
Mathematicians say "undefined" because they mean to say that $\frac{x}{0}$ is undefined within elementary algebra. However, they see $\frac{x}{0}$ as definable since they're not constricted to elementary algebra. | {
"domain": "sciencechatforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.97241471777321,
"lm_q1q2_score": 0.8596528101449367,
"lm_q2_score": 0.8840392832736084,
"openwebmath_perplexity": 1630.4931841615064,
"openwebmath_score": 0.8781976103782654,
"tags": null,
"url": "http://sciencechatforum.com/viewtopic.php?f=19&t=29316&p=303488&sid=7c166a4316456a6b426d7668e4965885"
} |
If we were to take a more abstract algebra approach, we could just state
${\text{Watson}}{\left(x\right)}{\equiv}{\frac{x}{0}}$.
And I know what you're thinking: how's that help, right? Well, simple: the ${\text{Watson}}{\left(x\right)}$ function doesn't return a normal number, but rather some new mathematical entity. This new mathematical entity obeys
$0{\times}{\text{Watson}}{\left(x\right)}=x$,
so it doesn't suffer from the automatic inconsistency that all elementary entities do in the post that I made above.
What other properties does this new mathematical entity have? I dunno, I'm too lazy to make them up right now. And, heck, it's named after you, so you make them up. You're allowed to make up whatever you want so long as you rigorously avoid contradicting yourself.
To sum this up, yeah, you're right that "undefinable" would make more sense to most folks since we usually think about math in the context of elementary algebra. However, since mathematicians don't see things that way, to them it's moreso "undefined" than "undefinable".
-----
Separately, I forgot to address your question about how undefined and indeterminate are somewhat the same. Well, in a way, they're actually exact opposites:
1. undefined: No elementary description fits.
2. indeterminate: The set of elementary descriptions which fit includes non-equivalent members.
However, these two terms are similar in that they refer to cases in which there isn't exactly one known set of equivalent elementary representations. | {
"domain": "sciencechatforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.97241471777321,
"lm_q1q2_score": 0.8596528101449367,
"lm_q2_score": 0.8840392832736084,
"openwebmath_perplexity": 1630.4931841615064,
"openwebmath_score": 0.8781976103782654,
"tags": null,
"url": "http://sciencechatforum.com/viewtopic.php?f=19&t=29316&p=303488&sid=7c166a4316456a6b426d7668e4965885"
} |
By which I mean:
1. Most elementary entities are defined by only one set of equivalent representations. Examples of representations for $k=2$:
• $k=0+2$
• $k=0+1+1$
• $k=3-1+1-3+2.5-0.5$
2. Undefined entities have no elementary representations. Examples of representations for $k={\frac{2}{0}}$:
• [None. $\frac{2}{0}$ can't even represent itself because it leads to a contradiction; you're literally not even allowed to write it! It's meaningless gibberish.].
3. Indeterminate entities have more than one possible set of equivalent representations. Examples of possible sets of equivalent representations for $k=\frac{0}{0}$:
• Set of representations in which $k=0$, including:
• $k=0$
• $k=0+0$
• $k=1-1$
• $k=0*0$
• $k=0.5-\frac{1}{2}$
• Set of representations in which $k=1$, including:
• $k=1$
• $k=1+0$
• $k=2-1$
• $k=1*1$
• $k=1.5-\frac{1}{2}$
• Set of representations in which $k=2$, including:
• $k=2$
• $k=1+1$
• $k=2-0$
• $k=1*2$
• $k=1.5+\frac{1}{2}$
• [Etc., since $k$ can be just about anything in this case.]
To sum this up, "undefined" and "indeterminate" are similar in that they both indicate that there's not exactly one set of equivalent elementary representations. However, "undefined" indicates zero sets while "indeterminate" indicates more than one set.
Natural ChemE
Forum Moderator
Posts: 2754
Joined: 28 Dec 2009
Braininvat, Darby liked this post
### Re: Is 0/0 undefined or indeterminate?
Exhaustively well said, but I think it might be easier for many to consider that undefined & indeterminate (or undefinable and indeterminable, as Watson put it) are somewhat similar only in a purely linguistic sense, whereas they are quite dissimilar in a purely mathematical sense ... kinda like 'alien' and 'illegal alien' are superficially similar linquistically, and yet are completely very different idiomatically. | {
"domain": "sciencechatforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.97241471777321,
"lm_q1q2_score": 0.8596528101449367,
"lm_q2_score": 0.8840392832736084,
"openwebmath_perplexity": 1630.4931841615064,
"openwebmath_score": 0.8781976103782654,
"tags": null,
"url": "http://sciencechatforum.com/viewtopic.php?f=19&t=29316&p=303488&sid=7c166a4316456a6b426d7668e4965885"
} |
Mathematics and linquistics share a lot of common ground, and indeed the former was birthed out of the latter, but it is contextually important to remember their differences and that they are not interchangeable - mathematics is more rigorously symbol oriented, whereas linguistics is generally more meaning oriented.
Math is a language unto itself.
Darby
Active Member
Posts: 1188
Joined: 14 Feb 2015
Location: Long Island, New York (USA)
### Re: Is 0/0 undefined or indeterminate?
Watson » August 17th, 2015, 5:58 pm wrote:I hope this isn't off topic, but undefined suggests it can be, but just hasn't yet been defined. Would not undefinable be more accurate, in the context of the OP?
Here you have been shown that no mathematician would ever use a term like "undefinable", because if something is not defined within a domain of mathematics then a good mathematician would rapidly create a new domain where the thing IS defined (such domain would probably start from that definition itself).
For example they would use your (algebraically undefined) ratio 0/0 to represent a discontinuity in the function x/|x|: for x>0 the function has constant value 1, and its limit for x--> 0 (from the right) also is 1; for x<0 the function has value -1 and that is also its limit for x-->0 (from the left). Since at zero the function instantaneously jumps from -1 to +1, its value cannot be determined there. Still, it is defined, as you can plot and integrate the function with no problems.
[Forget this second part - it only is an attempt at pretending I am also a "good mathematician", simply because I share that attitude: if you find something that does not fit into your model, then claim that that's only a part of your model, you do have a better, more complex one!]
neuro
Forum Moderator
Posts: 2635
Joined: 25 Jun 2010
Location: italy
### Re: Is 0/0 undefined or indeterminate? | {
"domain": "sciencechatforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.97241471777321,
"lm_q1q2_score": 0.8596528101449367,
"lm_q2_score": 0.8840392832736084,
"openwebmath_perplexity": 1630.4931841615064,
"openwebmath_score": 0.8781976103782654,
"tags": null,
"url": "http://sciencechatforum.com/viewtopic.php?f=19&t=29316&p=303488&sid=7c166a4316456a6b426d7668e4965885"
} |
Posts: 2635
Joined: 25 Jun 2010
Location: italy
### Re: Is 0/0 undefined or indeterminate?
if something is not defined within a domain of mathematics then a good mathematician would rapidly create a new domain where the thing IS defined (such domain would probably start from that definition itself).
Yeah you can do that, but it depends on the professor teaching it, and on the context where the expression appears during the course. If you have a series where the numerator approaches a finite value, say 2 , and the denominator is trending to zero
$\frac{2}{0}$
you can justifiably define a "new domain" where the real answer to the problem is $\infty$
Confusion arrises in introductory courses where the students are asked to instead provide the answer as "Does not exist".
The differentiation between "undefined" and "indeterminate" goes beyond pathological issues presented by Natural ChemE. In l'Hopital's Rule, the expression is required to be indeterminate. Otherwise, in expressions that are undefined, the theorem does not hold. (more importantly) you cannot use the technique to solve the limit.
hyksos
Active Member
Posts: 1294
Joined: 28 Nov 2014
### Re: Is 0/0 undefined or indeterminate?
hyksos,
That's a common misconception, but L'Hôpital's rule isn't meaningfully related to this topic.
L'Hôpital's rule is an alternative way to evaluate the limit of fractions. While it's usually more work than simpler methods, sometimes it's actually easier or/and provides a better result.
The commonly discussed case for L'Hôpital's rule is when just substituting in the approached value would result in either $\frac{0}{0}$ or $\frac{{\infty}}{{\infty}}$. The problem with both of these results is that they're not very descriptive, i.e. they're indeterminate. Math books often recommend considering L'Hôpital's rule when this happens since it might provide a determinate result.
Natural ChemE
Forum Moderator
Posts: 2754
Joined: 28 Dec 2009 | {
"domain": "sciencechatforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.97241471777321,
"lm_q1q2_score": 0.8596528101449367,
"lm_q2_score": 0.8840392832736084,
"openwebmath_perplexity": 1630.4931841615064,
"openwebmath_score": 0.8781976103782654,
"tags": null,
"url": "http://sciencechatforum.com/viewtopic.php?f=19&t=29316&p=303488&sid=7c166a4316456a6b426d7668e4965885"
} |
Posts: 2754
Joined: 28 Dec 2009
### Re: Is 0/0 undefined or indeterminate?
Natural ChemE.
L'Hôpital's rule is not just for $\frac{0}{0}$ and$\frac{ \infty }{ \infty }$ as you have claimed. The rule will work on any situation that is an indeterminate form. You have given two examples . There are many more, such as $0^0$ and several more others.
However, $\frac{2}{0}$ is NOT indeterminate! That's where the danger comes in. L'Hôpital's rule cannot be applied to situations that are "formally infinity", or in limits and series that genuinely diverge.
The main thing I wanted to communicate here is : yes, I agree with you that students are told to plug-and-chug L'Hôpital's rule, as you have described. However, dig in the literature a little deeper. L'Hôpital's rule is buttressed by two historical proofs and their corresponding theorems. This differentiation between indeterminate and undefined is very real in mathematics, and it has real consequences in theorems.
hyksos
Active Member
Posts: 1294
Joined: 28 Nov 2014
### Re: Is 0/0 undefined or indeterminate?
hksos,
Natural ChemE » August 31st, 2015, 7:55 pm wrote:The commonly discussed case for L'Hôpital's rule is when just substituting in the approached value would result in either $\frac{0}{0}$ or $\frac{{\infty}}{{\infty}}$.
hksos » September 12th, 2015, 4:17 pm wrote:L'Hôpital's rule is not just for $\frac{0}{0}$ and$\frac{ \infty }{ \infty }$ as you have claimed.
How the heck did you get from my statement to your representation of it?
There's a lot wrong with what you've said. Unless you can demonstrate some relevance, I'll just split 'em off to avoid polluting a good thread.
Natural ChemE
Forum Moderator
Posts: 2754
Joined: 28 Dec 2009
### Re: Is 0/0 undefined or indeterminate?
[quote="[url=http://www.sciencechatforum.com/viewtopic.php?p=286235#p286235]Natural ChemE » August 17th, 2015, 1:26 am[/url]"]krum, | {
"domain": "sciencechatforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.97241471777321,
"lm_q1q2_score": 0.8596528101449367,
"lm_q2_score": 0.8840392832736084,
"openwebmath_perplexity": 1630.4931841615064,
"openwebmath_score": 0.8781976103782654,
"tags": null,
"url": "http://sciencechatforum.com/viewtopic.php?f=19&t=29316&p=303488&sid=7c166a4316456a6b426d7668e4965885"
} |
What's $$\frac{x}{0}$$? Okay, let's ask that question mathematically by:[list=1]
[*]Call the definition $$k$$.
[*]Write that $$k=\frac{x}{0}$$.
[*]Solve for $$x$$:
[list]
[*]$$k=\frac{x}{0}$$;
[*]$$0{\times}k=0{\times}{\frac{x}{0}}$$
[*]$$0{\times}k=x$$
[*]$$x=0$$.[/list]
[*]Consider $$x=0$$.
[list]
[*]No definition $$k$$ would lead to a contradiction.
[*]So all possible definitions $$k$$ may be consistent.
[*]So we cannot determine a particular definition that $$k$$ must be.
[*]Call this inability to determine "indeterminate".[/list]
[*]Consider $$x{\neq}0$$.
[list]
[*]Any definition $$k$$ would lead to a contradiction.
[*]So there is no possible definition $$k$$.
[*]Call this inability to define "undefined".[/list][/list]
Therefore $$\frac{x}{0}$$ is:[list=i]
[*]indeterminate when $$x=0$$.
[*]undefined when $$x{\neq}0$$.[/list]
Historically some mathematicians have failed to rigorously consider branching cases when they had fractions. For example, if their proof contained $$\frac{x}{y}$$ at some intermediate step but was later removed (e.g. by multiplication by $$y$$), then it may be hard to tell, but the proof may be flawed through reliance on $$0{\times}{\frac{x}{0}}=x$$. This [url=https://en.wikipedia.org/wiki/Division_by_zero#Fallacies]hidden division-by-zero fallacy[/url] was the first major math fallacy that I was warned about.[/quote]
The bottom line is that 0/0 is not equal to any number and a/0 where a is not 0 is not any number.
Robert Kolker
Forum Neophyte
Posts: 1
Joined: 20 Jan 2016
### Re: Is 0/0 undefined or indeterminate?
It is indeterminate.
williamjohn
Forum Neophyte
Posts: 1
Joined: 20 May 2016
### Re: Is 0/0 undefined or indeterminate?
n/0 does not exist, for all n.
1. x/y defined as (the z such that y*z = x).
2. 0*z = 0, for all z.
n/0 = (the z: 0*z = n).
If ~(n=0) then there is no z such that 0*z = n. | {
"domain": "sciencechatforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.97241471777321,
"lm_q1q2_score": 0.8596528101449367,
"lm_q2_score": 0.8840392832736084,
"openwebmath_perplexity": 1630.4931841615064,
"openwebmath_score": 0.8781976103782654,
"tags": null,
"url": "http://sciencechatforum.com/viewtopic.php?f=19&t=29316&p=303488&sid=7c166a4316456a6b426d7668e4965885"
} |
n/0 = (the z: 0*z = n).
If ~(n=0) then there is no z such that 0*z = n.
e.g. (1/0) = (the z such that 0*z = 1).
But 2 asserts that all values of 0*z are equal to 0,
i.e. there is no value of z for which 0*z=1,
If there is no value of z such that 0*z=1 then
(the z such that 0*z = 1) does not exist.
(the z: 0*z = n) is not unique for all n > 0.
If n=0 then there is more than one value of z such that 0*z=0.
0*z=0 is true for all z.
i.e. there is no unique value of z such that 0*z = 0.
That is to say, n/0 does not exist for all values of n,
even though n/0 is defined.
Owen
Member
Posts: 80
Joined: 06 May 2007 | {
"domain": "sciencechatforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.97241471777321,
"lm_q1q2_score": 0.8596528101449367,
"lm_q2_score": 0.8840392832736084,
"openwebmath_perplexity": 1630.4931841615064,
"openwebmath_score": 0.8781976103782654,
"tags": null,
"url": "http://sciencechatforum.com/viewtopic.php?f=19&t=29316&p=303488&sid=7c166a4316456a6b426d7668e4965885"
} |
# How to pick $10$ people from $13$ such that at least $1$ is a woman
Thirteen people on a softball team show up for a game. Of the $13$ people who show up, $3$ are women. How many ways are there to choose $10$ players to take the field if at least one of these players must be a woman?
The given answer is calculated by summing the combination of $1$ woman + $9$ men, $2$ women + $8$ men, and $3$ women + $7$ men.
My question is, why can't we set this up as the sum $\binom{3}{1} + \binom{12}{9}$ - picking one of the three women first, then picking $9$ from the remaining $12$ men and women combined? The only requirement is that we have at least one woman, which is satisfied by $\binom{3}{1}$, and that leaves a pool of $12$ from which to pick the remaining $9$. The answer this way is close to the answer given, but it's $62$ short. I get that it's the "wrong" answer but I'm wondering why my thinking was wrong in setting it up this way. Thanks.
• You must multiply $C(3,1)$ with $C(12,9)$ because for every choice of a woman you have $C(12,9)$ choices for the other people. But this does not work either because you count arrangements multiple times. – Peter Jun 12 '16 at 22:15
Firstly you multiply, not add, if you are thinking of $\dbinom31$ and $\dbinom{12}{9}$
Secondly, this approach will over count. Suppose you chose Alicia , and then you chose $9$ from the remaining $12$, you would also have combos where Britney was first chosen, and Alicia was chosen from the $12$ group.
Thirdly, your book approach is correct, but unnecessarily tedious.
The best way is to compute
[All possible combos] - [All male combos ] $\;=\dbinom{13}{10} - \dbinom{10}{10}$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9796676448764483,
"lm_q1q2_score": 0.8596356303818424,
"lm_q2_score": 0.877476800298183,
"openwebmath_perplexity": 379.9378384899396,
"openwebmath_score": 0.7504289150238037,
"tags": null,
"url": "https://math.stackexchange.com/questions/1823736/how-to-pick-10-people-from-13-such-that-at-least-1-is-a-woman"
} |
• I get what you are saying, but the constraint is "at least one", so wouldn't having a combination where Alicia is picked first, followed by a combination where Britney is picked and then Alicia is picked, be perfectly valid? Because there is still "at least one" in each combination. Unless I misunderstood you. – Dave Jun 12 '16 at 22:51
• $A$ followed by $B$, and $B$ followed by $A$ are valid, but you are counting them twice (apart from other overcounts). Problems of the "at least one" type are best tackled using the complement. You can verify for yourself that the book method, and the above method give the same answer. – true blue anil Jun 13 '16 at 4:44
Among $13$ people, you can choose $10$ players in $$\binom {13}{10}$$ ways.
Now, suppose there is no woman in the squad, then you can choose that team in $\binom{10}{10}=1$ way.
So, number of ways of choosing a team with at least $1$ woman is $$\binom{13}3-1=286-1=285$$ ways.
When you are counting several countings(call it nested counting), you should apply the multiplication principle. Here counting ways of choosing $1$ woman and $9$ men, you are in some kind of this nested counting, where, for any $9$ men among the $10$, the woman can occur, so, have to multiply this.
And upon deriving each choices by multiplying, you add those terms. Because, in one counting, there is $1$ woman , in some other count, there are $2$, and in other, there are $3$, women.
So, all are separate cases, so add them, like $$\binom31 \binom{10}9+\binom 32\binom {10}8+\binom33\binom {10}7=285$$
• This is a better solution than the one on the supplied answer sheet. On the other hand, it doesn't explain why OP's method fails. – David K Jun 12 '16 at 22:18 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9796676448764483,
"lm_q1q2_score": 0.8596356303818424,
"lm_q2_score": 0.877476800298183,
"openwebmath_perplexity": 379.9378384899396,
"openwebmath_score": 0.7504289150238037,
"tags": null,
"url": "https://math.stackexchange.com/questions/1823736/how-to-pick-10-people-from-13-such-that-at-least-1-is-a-woman"
} |
# Convex hull of idempotent matrices
What is the convex hull of the set of $n\times n$ (potentially asymmetric) idempotent matrices?
Consider the set $S:=\{A\in\mathbb{R}^{n\times n}:A^2=A\}$, the set of idempotent matrices. Is there a simple description of the convex hull of $S$?
By convex hull, I mean: $C:=\{\sum_i a_i M_i:\sum_ia_i=1,a_i\geq0,M_i\in S\}$
• At the very least, we know that every matrix in the hull has a trace between $0$ and $n$. For all I know, we might get every matrix with trace between $0$ and $n$. – Omnomnomnom Oct 9 '15 at 18:34
• A condition on trace seems quite loose, however. Is there a smaller convex set containing the idempotent matrices? – Justin Solomon Oct 9 '15 at 18:42
• I really don't know. By the way, it looks like your question is about to be closed because it "needs more context". – Omnomnomnom Oct 9 '15 at 18:46
• I think the problem that people may have is that it looks like you didn't put enough work into answering the question yourself. – Omnomnomnom Oct 9 '15 at 18:52
• Sure, but the people of the site and the powers that be tend to expect some kind of partial work, though not necessarily a partial result. I happen to think your question is fine as is. – Omnomnomnom Oct 9 '15 at 18:55 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9796676472509722,
"lm_q1q2_score": 0.8596356167707372,
"lm_q2_score": 0.8774767842777551,
"openwebmath_perplexity": 142.51201729522447,
"openwebmath_score": 0.933679461479187,
"tags": null,
"url": "https://math.stackexchange.com/questions/1472277/convex-hull-of-idempotent-matrices"
} |
First I'll do the $2 \times 2$ case. Consider a $2 \times 2$ matrix of trace $1$:
$$A = \pmatrix{a & b\cr c & 1-a\cr}$$ For any real $r$, every point $(b,c)$ in the plane is a convex combination of two points on the hyperbola $x y = -r$ (the two axes in the case $r=0$). Thus $A$ is a convex combination of two idempotent matrices of the form $$\pmatrix{ a & x\cr y & 1-a\cr}$$ where $xy = a-a^2$. The only idempotent matrix of trace $0$ is $0$, and the only idempotent matrix of trace $2$ is $I$. Any matrix of trace strictly between $0$ and $1$ is a convex combination of $0$ and a matrix of trace $1$, while any matrix of trace strictly between $1$ and $2$ is a convex combination of $I$ and a matrix of trace $1$. Thus the convex hull in the $2 \times 2$ case consists of $0$, $I$ and all $A$ with $0 < \text{tr}(A) < 2$.
This leads to a solution in the $n \times n$ case for any $n > 2$. We have idempotent matrices of trace $1$ with the form $$\pmatrix{a & x & 0 & \ldots & 0\cr y & 1-a & 0 & \ldots & 0\cr 0 & 0 & 0 & \ldots & 0\cr \ldots & \ldots & \ldots & \ldots & \ldots\cr 0 & 0 & 0 & \ldots & 0\cr}$$ where $xy = a - a^2$, and convex combinations of these give all $n \times n$ matrices of trace $1$ that are $0$ outside the top left $2 \times 2$ submatrix. By permuting indices, we get all $n \times n$ matrices of trace $1$ that are $0$ outside some principal $2 \times 2$ submatrix. Convex combinations of these give us all $n \times n$ matrices of trace $1$. Now any $n \times n$ matrix of trace strictly between $0$ and $1$ is a convex combination of $0$ and a matrix of trace $1$, while any matrix of trace strictly between $1$ and $n$ is a convex combination of $I$ and a matrix of trace $1$. Thus we find that the convex hull consists of $0$, $I$ and all $A$ with $0 < \text{tr}(A) < n$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9796676472509722,
"lm_q1q2_score": 0.8596356167707372,
"lm_q2_score": 0.8774767842777551,
"openwebmath_perplexity": 142.51201729522447,
"openwebmath_score": 0.933679461479187,
"tags": null,
"url": "https://math.stackexchange.com/questions/1472277/convex-hull-of-idempotent-matrices"
} |
• so do we know that there won't be any matrices with trace $2$ besides $I$? – Omnomnomnom Oct 9 '15 at 19:14
• Also, what about matrices with trace $0$? – Omnomnomnom Oct 9 '15 at 19:17
• An idempotent matrix of trace $k$ is a projection on a subspace of dimension $k$. The only subspace of dimension $2$ is the whole space, and a projection on that is the identity. The only subspace of dimension $0$ is $\{0\}$, and a projection on that is $0$. – Robert Israel Oct 9 '15 at 19:25
• I see, there is a unique idempotent element with trace $2$ (or $0$), and so the convex hull must also have a unique such element. – Omnomnomnom Oct 9 '15 at 19:32
• Wow! Thanks to both of you for your help -- it's much appreciated. I'm surprised the convex hull is so big. – Justin Solomon Oct 9 '15 at 19:50
The result can be generalized by induction. Let $C_n$ denote the convex hull over $n \times n$ matrices.
First of all, note that if $A \in C_{n-1}$ and $x \in \Bbb R^n$, then the elements $$\pmatrix{A&0\\0&0_{1\times 1}},E_x := \pmatrix{I_{n-1}&x\\0&0}$$ are all in $C_n$.
Next, note that $C_n$ is closed under similarity (anything similar to an element of $C_n$ is in $C_n$).
Now, take any upper triangular $A \in C_{n-1}$. That is, $$A = \pmatrix{\lambda_1 & * \cdots \\ &\ddots\\ && \lambda_{n-1}}$$ with $\sum\lambda_i \in (0,n-1)$. Note that any convex combination of $A$ and $I$ is in $C_n$. Also, any convex combination of a matrix $C_n$ with $E_x$ is in $C_n$.
Conclude that $C_n$ contains every upper triangular matrix with trace (strictly) between $0$ and $n$ and a non-negative bottom-right entry. Because every matrix is upper-triangularizable, conclude that $C_n$ contains every matrix with trace strictly between $0$ and $n$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9796676472509722,
"lm_q1q2_score": 0.8596356167707372,
"lm_q2_score": 0.8774767842777551,
"openwebmath_perplexity": 142.51201729522447,
"openwebmath_score": 0.933679461479187,
"tags": null,
"url": "https://math.stackexchange.com/questions/1472277/convex-hull-of-idempotent-matrices"
} |
A very interesting question: intersection point of $x^y=y^x$
I've been investigating the Cartesian graph of $$x^y=y^x$$. Obviously, part of the graph is comprised of the line $$y=x$$ but there is also a curve that is symmetrical about the line $$y=x$$. (We can prove this symmetry by noting that the function $$x^y=y^x$$ is self-inverse; all self-inverse functions are symmetrical about the line $$y=x$$.)
An image of the graph is shown below:
I decided to find the intersection point and arrived at an intriguing result: the intersection point between the two curves is at $$(e,e)$$.
The following is my method: If the gradient of the line $$y=x$$ is $$1$$, the gradient of the curve at the intersection point must be $$-1$$ as it's a normal to the line (as it's symmetrical about the line). This means that at that point $$\frac{dy}{dx}=-1$$. Now to find $$\frac{dy}{dx}$$.
We have $$x^y=y^x$$. I then used a very powerful tehnique for differentiating these sorts of functions. We know that eg $$x^y=e^{\ln{x^y}}=e^{y\ln{x}}$$ and $$\frac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}$$ Applying it to our function $$x^y=y^x$$ and using implicit differentiation and the product rule gives us: $$(\frac{dy}{dx}\times \ln x +\frac{y}{x})x^y=(\ln y +\frac{dy}{dx}\times \frac{x}{y})y^x$$ So $$\frac{dy}{dx}x^y\ln x +yx^{y-1}=\frac{dy}{dx}xy^{x-1}+y^x \ln y$$ Extensively rearranging gives: $$\frac{dy}{dx}=\frac{y^x \ln y -yx^{y-1}}{x^y \ln x -xy^{x-1}}$$ Let $$\frac{dy}{dx}=-1$$: $$y^x \ln y -yx^{y-1}=xy^{x-1}-x^y\ln x$$ But we know $$x=y$$ since we are at the intersection point with the line $$y=x$$: $$x^x \ln x -x^x=x^x-x^x\ln x$$ So $$2x^x \ln x -2x^x=0$$ $$2x^x(\ln x -1)=0$$ This means either $$2x^x=0$$ or $$\ln x -1=0$$ but we know $$x^x$$ is always greater than $$0$$ so $$\ln x =1$$, leaving us with: $$x=y=e$$ So I have my answer, but is there any other method of getting it? I have heard there is. Any help will be very welcome. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9914225143328517,
"lm_q1q2_score": 0.8595987610775709,
"lm_q2_score": 0.8670357477770336,
"openwebmath_perplexity": 189.3965862014512,
"openwebmath_score": 0.9445435404777527,
"tags": null,
"url": "https://math.stackexchange.com/questions/3724980/a-very-interesting-question-intersection-point-of-xy-yx"
} |
• The solution to $y^x=x^y$ can be expressed in terms of Lambert's W function as $$y=-xW\left(-\log(x)/x\right)/\log(x)$$Lambert's W function has been studied extensively. [Here](en.wikipedia.org/wiki/Lambert_W_function) is a link to a Wikipedia article that might be of interest. Note in particular that $W(-1/e)=-1$. Jun 18 '20 at 14:42
• Does this answer your question? Where does the curve $x^y = y^x$ intersect itself? Jun 18 '20 at 15:13
• Another possible duplicate: Given that $x^y=y^x$, what could $x$ and $y$ be?. Jun 18 '20 at 15:14
• Does this answer your question? Given that $x^y=y^x$, what could $x$ and $y$ be? Jun 18 '20 at 16:11
• Only the question titled 'Where does the curve $x^y=y^x$ intersect itself?' is helpful, but it still isn't very detailed when explaining the limits.Thanks for the suggestions though! Jun 18 '20 at 18:15
A Simple Approach
The simplest approach that I have found is to look at the intersections of $$y=tx\qquad\text{and}\qquad x^y=y^x\tag1$$ That is, \begin{align} x^{tx}&=(tx)^x\tag{2a}\\[3pt] x^t&=tx\tag{2b}\\[3pt] x^{t-1}&=t\tag{2c}\\ x&=t^{\frac1{t-1}}\tag{2d}\\ y&=t^{\frac t{t-1}}\tag{2e} \end{align} Explanation:
$$\text{(2a)}$$: $$x^y=y^x$$
$$\text{(2b)}$$: raise to the $$\frac1x$$ power
$$\text{(2c)}$$: divide by $$x$$
$$\text{(2d)}$$: raise to the $$\frac1{t-1}$$ power
$$\text{(2e)}$$: $$y=tx$$
Now, if we wish to find where $$x=y$$, let $$t\to1$$. That is, \begin{align} x &=\lim_{t\to1}t^{\frac1{t-1}}\tag{3a}\\ &=\lim_{n\to\infty}\left(1+\frac1n\right)^n\tag{3b}\\[6pt] &=e\tag{3c} \end{align} Explanation:
$$\text{(3a)}$$: $$x=y$$ when $$t=1$$
$$\text{(3b)}$$: $$t=1+\frac1n$$
$$\text{(3c)}$$: evaluate the limit
Further Musings
We can also compute \begin{align} y &=\lim_{t\to1}t^{\frac t{t-1}}\tag{4a}\\ &=\lim_{n\to\infty}\left(1+\frac1n\right)^{n+1}\tag{4b}\\[6pt] &=e\tag{4c} \end{align} Explanation:
$$\text{(4a)}$$: $$x=y$$ when $$t=1$$
$$\text{(4b)}$$: $$t=1+\frac1n$$
$$\text{(4c)}$$: evaluate the limit | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9914225143328517,
"lm_q1q2_score": 0.8595987610775709,
"lm_q2_score": 0.8670357477770336,
"openwebmath_perplexity": 189.3965862014512,
"openwebmath_score": 0.9445435404777527,
"tags": null,
"url": "https://math.stackexchange.com/questions/3724980/a-very-interesting-question-intersection-point-of-xy-yx"
} |
Using the results from this answer, we see that $$\left(1+\frac1n\right)^n$$ is increasing and $$\left(1+\frac1n\right)^{n+1}$$ is decreasing, which means that $$\left(1+\frac1n\right)^n\le e\le\left(1+\frac1n\right)^{n+1}$$.
Thus, as $$t\to1^+$$, $$(4)$$ and $$(5)$$ show that $$x\to e^-$$ and $$y\to e^+$$.
Furthermore, if we substitute $$t\mapsto1/t$$, we get \begin{align} t^{\frac1{t-1}} &\mapsto(1/t)^{\frac1{1/t-1}}\tag{5a}\\ &=t^{\frac t{t-1}}\tag{5b} \end{align} and \begin{align} t^{\frac t{t-1}} &\mapsto(1/t)^{\frac{1/t}{1/t-1}}\tag{6a}\\ &=t^{\frac1{t-1}}\tag{6b} \end{align} That is, substituting $$t\mapsto1/t$$ swaps $$x$$ and $$y$$.
This means that, as $$t\to1^-$$, $$x\to e^+$$ and $$y\to e^-$$.
Relation to the Graph
Here is where these points sit on the graph as $$t\to1^+$$:
• Incredibly clear, thank you!!! Jun 23 '20 at 14:33
• @robjohn Your answers always have fantastic graphics and explanations. Nicely done, +1 from me. Jul 7 '20 at 5:35
Not sure to understand well the question but if it provides any help, I think there is something missing in your analysis.
Your equation can be seen as $$g(x,y) = x^{y} = y^{x} = f(x,y)$$. You are looking for intersecting points or surfaces (x,y) in $$\mathbb{R}^{3}$$, since both $$f(\cdot,\cdot)$$ and $$g(\cdot,\cdot)$$ are 2D scalar fields. Otherwise, if you are thinking the expression $$x^{y} = y^{x}$$ as the equation of a curve in $$\mathbb{R}$$ (somehow you figure out to express $$y = f(x)$$ or $$x = f(x)$$, it doesn't make sense to talk about an intercept (or interception set) since you must provide another curve or surface.
The same applies with the fields $$g$$ and $$f$$. You must provide another equation since you have two unknowns.
The answer you got $$(x,y) = (e,e)$$ is the trivial answer, and works for any real $$k \in \mathbb{R}$$: $$(x,y) = (k,k)$$.
I don't know if this is what you are looking for, but I would go like this. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9914225143328517,
"lm_q1q2_score": 0.8595987610775709,
"lm_q2_score": 0.8670357477770336,
"openwebmath_perplexity": 189.3965862014512,
"openwebmath_score": 0.9445435404777527,
"tags": null,
"url": "https://math.stackexchange.com/questions/3724980/a-very-interesting-question-intersection-point-of-xy-yx"
} |
I don't know if this is what you are looking for, but I would go like this.
First rewrite the equation : $$x^y=y^x\Leftrightarrow y\ln(x)=x\ln(y)\Leftrightarrow \frac{\ln(x)}{x}=\frac{\ln(y)}{y}$$ Now consider the function $$f(x)=\frac{\ln(x)}{x}$$ defined on $$(0,+\infty)$$ and whose derivative is $$f'(x)=\frac{1-\ln(x)}{x^2}$$. This shows that $$f$$ induces two bijections $$f_1:(0,e]\to (-\infty,\frac{1}{e}]$$ and $$f_2:[e,+\infty)\to[\frac{1}{e},0)$$.
Now the locus $$f(x)=f(y)$$ can be divided in two curves, namely $$y=f_1^{-1}(f(x))$$ and $$y=f_2^{-1}(f(x))$$. (Note however that these two curves are not the obvious ones). And we are looking their intersection. Since the ranges of $$f_1^{-1}$$ and $$f_2^{-1}$$ are respectively $$(0,e]$$ and $$[e,+\infty)$$, the only possible value of $$y$$ for the intersection is $$y=e$$ which happen when $$x=e$$.
• Actually, this method shows that for a function $f$ which has a single change of variation (hence a unique local extrema $x_0$), then $f(x)=f(y)$ has two branches whose intersection is exactly $(x_0,x_0)$. Jun 18 '20 at 15:46
This is not exactly an answer, but part of your reasoning was a bit hand-wavy and I want to make it fully rigorous.
Your reasoning about the slopes of the two parts of the plot is intuitively correct, but needs proof. So it is clear that the slope of the first part of the plot is $$1$$ since this encompasses the solutions $$x^x=x^x$$, i.e, the identity map which has slope $$1$$ by definition. However, showing that the other branch of the plot has slope $$-1$$ is a bit more difficult. We need to show that
1: The relation $$y^x-x^y=0$$ is its own inverse
2: If a relation is its own inverse, its slope at a point $$(x,y)$$ is $$-1$$ if $$x=y$$, with the exception of the identity map. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9914225143328517,
"lm_q1q2_score": 0.8595987610775709,
"lm_q2_score": 0.8670357477770336,
"openwebmath_perplexity": 189.3965862014512,
"openwebmath_score": 0.9445435404777527,
"tags": null,
"url": "https://math.stackexchange.com/questions/3724980/a-very-interesting-question-intersection-point-of-xy-yx"
} |
The first point is rather easy and I'll leave it for you to do yourself. The second part, however, requires a bit of care. Any two variable relation can be summarized by the general equation $$f(x,y)=0$$ Well, not quite any, but I hope you get the idea. For example, the equation of a circle: $$x^2+y^2-r^2=0$$ We want to use information about $$f$$ to deduce $$\frac{\mathrm{d}y}{\mathrm{d}x}$$. Using a bit of knowledge from multivariable calculus, we know that the total derivative of a two-variable function $$f$$ in terms of an arbitrary variable $$u$$ is $$\frac{\mathrm{d} f}{\mathrm{d} u}=\frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}u}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}u}$$ In the special case that $$u=x$$, $$\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x}$$ And similarly $$\frac{\mathrm{d} f}{\mathrm{d} y}=\frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}y}+\frac{\partial f}{\partial y}$$ If the relation is its own inverse, then $$f(x,y)=f(y,x)=0$$ Thus $$\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d} f}{\mathrm{d} y}=\frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}y}+\frac{\partial f}{\partial y}$$ So $$\frac{\partial f}{\partial x}\left(1-\frac{\mathrm{d}x}{\mathrm{d}y}\right)=\frac{\partial f}{\partial y}\left(1-\frac{\mathrm{d}y}{\mathrm{d}x}\right)$$ But again, since $$f(x,y)=f(y,x)=0$$, $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}$$ if $$x=y$$, thus $$1-\frac{\mathrm{d}x}{\mathrm{d}y}=1-\frac{\mathrm{d}y}{\mathrm{d}x}$$ $$\frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}}=\frac{\mathrm{d}y}{\mathrm{d}x}$$ Therefore $$\frac{\mathrm{d}y}{\mathrm{d}x}=\pm 1$$. Excluding the only self-inverse relation with slope $$1$$, the identity map, we conclude $$\frac{\mathrm{d}y}{\mathrm{d}x}\big|_{(x,x)}=-1.$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9914225143328517,
"lm_q1q2_score": 0.8595987610775709,
"lm_q2_score": 0.8670357477770336,
"openwebmath_perplexity": 189.3965862014512,
"openwebmath_score": 0.9445435404777527,
"tags": null,
"url": "https://math.stackexchange.com/questions/3724980/a-very-interesting-question-intersection-point-of-xy-yx"
} |
Following this, the rest of your reasoning is correct. Others have stated that this problem can be tackled using the product log, so I can't really think of another way of going about this problem. Good question though!
• Thanks for the detailed answer; I realized I didn't have ironclad proof for the second point, but that explanation is very helpful and welcome! Thanks a s well for the final compliment :) Jun 18 '20 at 18:05
We are looking for solutions to $$\exp(y\log x) = \exp(x\log y) \, .$$ Taking logs of both sides and rearranging, this equation becomes $$\frac{x}{\log x}=\frac{y}{\log y} \, .$$ Consider the graph of $$f(z)=z/\log z$$:
Its unusual shape can be explained by a few observations:
• For $$0, $$\log z$$ is negative, so $$f(z)$$ is negative.
• As $$z\to1$$, $$\log z \to 0$$, and so the graph has an asymptote at $$z=1$$.
• For $$z>1$$, $$\log z$$ is positive, so $$f(z)$$ is positive.
• $$f(z)$$ is large when $$\log z$$ is small, or when $$z$$ is very large.
It is the final observation that gives us a hint as to why $$f(z)$$ has a local minimum. We can study the behaviour of $$f(z)$$ more closely by considering $$f'(z) = \frac{\log z - 1}{\log^2z} \, .$$ For $$1, $$f'(z)$$ is negative; at $$z=e$$, $$f'(z)$$ is zero; and for $$z>e$$, $$f'(z)$$ is positive.
Now that we have proven that $$f(z)$$ has a local minimum at $$(e,e)$$, we can begin to examine your question more closely. Let $$(x,\log x)$$ and $$(y,\log y)$$ be arbitrary points that lies on the graph. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9914225143328517,
"lm_q1q2_score": 0.8595987610775709,
"lm_q2_score": 0.8670357477770336,
"openwebmath_perplexity": 189.3965862014512,
"openwebmath_score": 0.9445435404777527,
"tags": null,
"url": "https://math.stackexchange.com/questions/3724980/a-very-interesting-question-intersection-point-of-xy-yx"
} |
• Suppose $$0. Then, because $$f(z)$$ is one-to-one on the interval $$(0,1)$$, the only time when $$f(x)=f(y)$$ is when $$x=y$$.
• If $$x>1$$, then things are more complicated. Clearly, if $$x=y$$, then we still have $$f(x)=f(y)$$. However, if $$1, then there is a unique number $$y\in(e,\infty)$$ such that $$f(x)=f(y)$$. This is obviously still the case if the roles of $$x$$ and $$y$$ are interchanged.
• If $$x$$ is slightly smaller than $$e$$, then $$y$$ is either equal to $$x$$, or is only slightly larger than $$e$$. This means that the two sections of the graph get very close to each other as $$x$$ tends to $$e$$.
• In the limit, when $$x=e$$, the two parts of the graph coincide.
The equation is equivalent to
$$\frac{\log x}x=\frac{\log y}y.$$
The function
$$\frac{\log x}x$$
has a maximum at $$x=e$$ and for $$x>0$$
$$\frac{\log x}x=\frac{\log y}y$$ has two solutions in $$x$$.
Using Lambert's function,
$$-\frac1x\log\frac1x=\frac{\log y}y$$ and
$$e^{\log\frac1x}\log\frac1x=-\frac{\log y}y$$
and
$$\log\frac1x=W\left(-\frac{\log y}y\right).$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9914225143328517,
"lm_q1q2_score": 0.8595987610775709,
"lm_q2_score": 0.8670357477770336,
"openwebmath_perplexity": 189.3965862014512,
"openwebmath_score": 0.9445435404777527,
"tags": null,
"url": "https://math.stackexchange.com/questions/3724980/a-very-interesting-question-intersection-point-of-xy-yx"
} |
# A tangent to the ellipse meets the $x$ and $y$ axes . If $O$ is the origin, find the minimum area of triangle $AOB$.
The question is that : A tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the $x$ and $y$ axes respectively at $A$ and $B$. If $O$ is the origin, find the minimum area of triangle $AOB$.
What I have attempted:
Because the equation of an ellipse is $$\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$$
We have that the $x$ intercept is $x=\frac{a^2}{x_1}$ and $y$ intercept is $y=\frac{b^2}{y_1}$
Hence the area of a triangle $(AOB)$ is given by $A=\frac{a^2b^2}{2x_1y_1}$ now I am stuck, how should I proceed?
• What is the derivative $y'$? Aug 4, 2016 at 3:41
• Notice affine transformation preserves tangency and ratio of areas, you can use an affine transform to map the ellipse to the unit circle. The smallest triangle $AOB$ will be mapped to a right angled isoceles triangle of side $\sqrt{2}$ and area $1$. This means the smallest triangle $AOB$ has area $ab$. Aug 4, 2016 at 4:00
$A = \frac {a^2 b^2}{2xy}$ constrained by $\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$
$\frac{dA}{dx} = \frac {-2a^2b^2 (y + x y')}{(2xy)^2} = 0\\ y + x y' = 0\\ y' = -\frac yx$
Differentiating the constraint.
$\frac x{a^2} + \frac {y y'}{b^2} = 0\\ y' = -\frac {b^2x}{a^2y}\\ \frac {b^2x}{a^2y} = \frac yx\\ b^2 x^2 = a^2 y^2$
Again from the constraint:
$b^2 x^2 + a^2 y^2 = a^2 b^2\\ 2a^2 y^2 = a^2 b^2\\ y^2 = \frac {b^2}{2}\\ y = \frac b{\sqrt {2}}\\ 2xy = ab\\ A = \frac {a^2b^2}{2xy} = ab$
alternate
$x = a \cos t\\ y = b \sin t\\ A = \frac {ab}{2} \csc t \sec t\\ \frac{dA}{dt} = \frac {ab}{2} \sec t \csc t ( -\cot t + \tan t) = 0\\ \cot t = \tan t\\ t = \frac {\pi}{4}\\ A = ab$
That is much nicer. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9893474913588877,
"lm_q1q2_score": 0.85957158981653,
"lm_q2_score": 0.8688267745399466,
"openwebmath_perplexity": 104.9264473499854,
"openwebmath_score": 0.9090828895568848,
"tags": null,
"url": "https://math.stackexchange.com/questions/1881826/a-tangent-to-the-ellipse-meets-the-x-and-y-axes-if-o-is-the-origin-find"
} |
That is much nicer.
With the change of coordinates $x=au, y=bv$ the problem boils down to the same problem where the original ellipse is replaced by a unit circle. In such a case it is trivial that the minimum area of $AOB$ is $1$. Since the determinant of the Jacobian of the linear map $x=au, y=bv$ is $ab$, the answer to the original question is $\color{red}{ab}$, plain and simple. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9893474913588877,
"lm_q1q2_score": 0.85957158981653,
"lm_q2_score": 0.8688267745399466,
"openwebmath_perplexity": 104.9264473499854,
"openwebmath_score": 0.9090828895568848,
"tags": null,
"url": "https://math.stackexchange.com/questions/1881826/a-tangent-to-the-ellipse-meets-the-x-and-y-axes-if-o-is-the-origin-find"
} |
unit:PlanckTime
Type
Description
In physics, the Planck time, denoted by $$t_P$$, is the unit of time in the system of natural units known as Planck units. It is the time required for light to travel, in a vacuum, a distance of 1 Planck length. The unit is named after Max Planck, who was the first to propose it. $$\\ t_P \equiv \sqrt{\frac{\hbar G}{c^5}} \approx 5.39106(32) \times 10^{-44} s$$ where, $$c$$ is the speed of light in a vacuum, $$\hbar$$ is the reduced Planck's constant (defined as $$\hbar = \frac{h}{2 \pi}$$ and $$G$$ is the gravitational constant. The two digits between parentheses denote the standard error of the estimated value.
Properties
0.0000000000000000000000000000000000000000000539124
$$t_P$$
Annotations
In physics, the Planck time, denoted by $$t_P$$, is the unit of time in the system of natural units known as Planck units. It is the time required for light to travel, in a vacuum, a distance of 1 Planck length. The unit is named after Max Planck, who was the first to propose it. $$\\ t_P \equiv \sqrt{\frac{\hbar G}{c^5}} \approx 5.39106(32) \times 10^{-44} s$$ where, $$c$$ is the speed of light in a vacuum, $$\hbar$$ is the reduced Planck's constant (defined as $$\hbar = \frac{h}{2 \pi}$$ and $$G$$ is the gravitational constant. The two digits between parentheses denote the standard error of the estimated value.
Planck Time(en)
Generated 2022-11-28T15:45:48.023-05:00 by lmdoc version 1.1 with TopBraid SPARQL Web Pages (SWP) | {
"domain": "qudt.org",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9893474907307123,
"lm_q1q2_score": 0.8595715842303123,
"lm_q2_score": 0.868826769445233,
"openwebmath_perplexity": 179.52588681523468,
"openwebmath_score": 0.8906638622283936,
"tags": null,
"url": "https://qudt.org/vocab/unit/PlanckTime"
} |
# Generalized eigenvector for product of commuting matrices
Suppose $$A,B$$ are commuting invertible matrices with a common generalized eigenvector $$v$$ with eigenvalues $$a,b$$ respectively. That is, suppose there exist positive integers $$K,L$$ such that $$(A-aI)^K v= 0$$ and $$(B-bI)^Lv=0$$. Is it true that there exists a positive integer $$M$$ such that $$(AB-abI)^Mv = 0$$?
A bit of context: I came across this while thinking about the analogous version of this for eigenvalues (when $$K=1$$ and $$L=1$$). This is easy to show. In fact, the above statement is also easy to show if only one of $$K=1$$ or $$L=1$$ is true (that is, $$v$$ is an eigenvector for one matrix and a generalized eigenvector for the other).
Proof: WLOG suppose $$L=1$$. Then we have \begin{align} (AB-abI)^Kv &= \sum_{j=0}^K {K \choose j} A^{K-j}B^{K-j}a^jb^jv\\ &= \sum_{j=0}^K {K \choose j} A^{K-j}a^jb^Kv\\ &= b^K(A-aI)^Kv\\ &= 0 \end{align}
A natural question: is it true if $$v$$ is a generalized eigenvector for both matrices? | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9893474879039229,
"lm_q1q2_score": 0.8595715800941748,
"lm_q2_score": 0.8688267677469952,
"openwebmath_perplexity": 124.23096803506014,
"openwebmath_score": 0.9855372905731201,
"tags": null,
"url": "https://math.stackexchange.com/questions/4180075/generalized-eigenvector-for-product-of-commuting-matrices"
} |
A natural question: is it true if $$v$$ is a generalized eigenvector for both matrices?
• Welcome! Note that askers are expected to provide context for their questions, as is explained here. For example, it would be helpful if you could edit your questions to address some of the following. What motivated this problem, or where did you encounter it? What are your thoughts on the problem? What have you tried so far? – Ben Grossmann Jun 22 at 16:20
• A potentially helpful way to reframe the problem: let $P = A - aI$ and $Q = B - bI$. We could equivalently ask: if $PQ = QP$ and $K,L$ are such that $P^Kv = Q^Lv = 0$, then does there necessarily exist an $M$ such that $$([P + aI][Q + bI] - ab I)^Mv = 0?$$ – Ben Grossmann Jun 22 at 16:25
• The answer is yes: the statement is true. Once you provide some context, I can leave a proof as an answer. – Ben Grossmann Jun 22 at 16:46
• Just added some context. Thanks for pointing the rules out btw, I'm a new poster here. – Underbounded Jun 22 at 16:47
• I figured as much, thanks for obliging. – Ben Grossmann Jun 22 at 16:48
Here is a more abstract perspective. Let $$V$$ be the smallest subspace of our vector space which contains $$v$$ and is closed under the actions of $$A$$ and $$B$$. Let $$R$$ be the subring of the endomorphism ring of $$V$$ generated by the scalar matrices together with $$A$$ and $$B$$. By assumption, this $$R$$ is commutative, and $$(A-a)^K$$ and $$(B-b)^L$$ are equal to $$0$$ in this ring (since they annihilate $$v$$ and $$v$$ generates $$V$$ under the action of $$R$$). The question then becomes simply:
Let $$R$$ be a commutative ring and $$A,B,a,b\in R$$. Suppose $$A-a$$ and $$B-b$$ are nilpotent. Then must $$AB-ab$$ be nilpotent?
But now the answer is very easy using the fact that the set $$N$$ of nilpotent elements of $$R$$ is an ideal. By assumption, $$A=a$$ and $$B=b$$ in the quotient ring $$R/N$$. Thus $$AB=ab$$ in that quotient ring. That is, $$AB-ab\in N$$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9893474879039229,
"lm_q1q2_score": 0.8595715800941748,
"lm_q2_score": 0.8688267677469952,
"openwebmath_perplexity": 124.23096803506014,
"openwebmath_score": 0.9855372905731201,
"tags": null,
"url": "https://math.stackexchange.com/questions/4180075/generalized-eigenvector-for-product-of-commuting-matrices"
} |
• This is a really neat approach! – Underbounded Jun 22 at 21:23
Yes. We can proceed as follows.
Begin by reframing the question: if $$PQ = QP$$ and $$K,L$$ are such that $$P^Kv = Q^Lv = 0$$ and if $$a,b$$ are arbitrary scalars, then does there necessarily exist an $$M$$ such that $$([P + aI][Q + bI] - ab I)^Mv = 0?$$ Note that $$[P + aI][Q + bI] - ab I = PQ + bP + aQ.$$ Now, take $$m = \max\{K,L\}$$ and $$M = 2m-1$$; note that $$P^mv = Q^mv = 0$$. Apply the multinomial theorem: $$(PQ + bP + aQ)^Mv = \sum_{k_1 + k_2 + k_3 = M} \binom{M}{k_1,k_2,k_3} a^{k_3}b^{k_2}P^{k_1+k_2}Q^{k_1+k_3}v \\= \sum_{k_1 + k_2 + k_3 = M} \binom{M}{k_1,k_2,k_3} a^{k_3}b^{k_2} Q^{k_1+k_3}P^{k_1+k_2}v.$$ Now, for all triples of non-negative integers $$k_1,k_2,k_3$$ with $$k_1 + k_2 + k_3 = 2m-1$$, we see that $$k_1 + k_2$$ and $$k_1 + k_3$$ are non-negative integers with sum greater than or equal to $$2m-1$$. It follows that either $$k_1 + k_2 \geq m$$ or $$k_1 + k_3 \geq m$$. In either case, we find that $$P^{k_1+k_2}[Q^{k_1 + k_3}v] = Q^{k_1+k_3}[P^{k_1+k_2}v] = 0.$$ Thus, the above expansion of $$(PQ + bP + aQ)^Mv$$ must be equal to zero.
• Binomial theorem suffices if you write $AB-abI$ as $(A-aI)B+a(B-bI)$. This also lowers the value of $m$ to $K+L-1$. – user1551 Jun 22 at 17:45
• @user1551 That's neat! Actually, I realize now that $M = K +L - 1$ suffices for me as well; we only need $k_1 + k_2 \geq K$ or $k_1 + k_3 \geq L$ – Ben Grossmann Jun 22 at 17:57 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9893474879039229,
"lm_q1q2_score": 0.8595715800941748,
"lm_q2_score": 0.8688267677469952,
"openwebmath_perplexity": 124.23096803506014,
"openwebmath_score": 0.9855372905731201,
"tags": null,
"url": "https://math.stackexchange.com/questions/4180075/generalized-eigenvector-for-product-of-commuting-matrices"
} |
# What do the lines in a ContourPlot Mean?
If you plot a two variable function like $x^2+y^2$, you get a series of circles with lines and colored areas. My question is, what do the lines and the colored areas represent?
• Try this experiment: ContourPlot[Cos[x] + Cos[y], {x, 0, 4 Pi}, {y, 0, 4 Pi}], followed by Plot3D[Cos[x] + Cos[y], {x, 0, 4 Pi}, {y, 0, 4 Pi}] and compare the two. Now try ContourPlot[x^2 + y^2, {x, -4, 4 }, {y, -4, 4 }] and Plot3D[x^2 + y^2, {x, -4, 4 }, {y, -4, 4 }]. Got it? ContourPlot by default generates colorized grayscale output, in which larger values are shown lighter. You may also like to read: en.wikipedia.org/wiki/Contour_line and itl.nist.gov/div898/handbook/eda/section3/contour.htm – Moo Apr 23 '16 at 12:50
• @Moo, OK, so now I know why there are areas of different colors but what is the meaning if the lines bounding each colored area? Do they represent something? – MrDi Apr 23 '16 at 12:56
• It's explained well here: en.wikipedia.org/wiki/Contour_line The value of the function doesn't change as you move along a contour. – Szabolcs Apr 23 '16 at 13:10
• @MrDi: There needs to be a way to select peaks and valleys and to bound them somehow, so they use some algorithm that picks those areas. I think you can find some details on the Wiki link I provided. it sounds like you get it, think of it as trying to collapse a 3D image onto a 2D surface. These diagrams can be extremely useful. The lines are just separating areas where something is changing based on some algorithm. – Moo Apr 23 '16 at 13:13
• @Moo, OK, thanks for help. – MrDi Apr 23 '16 at 13:14
It might be easier for you to see the correspondence between a contour plot and a surface plot if you plot them side-by-side like this (code adapted from here):
peaks[x_, y_] := 3 (1 - x)^2 Exp[-x^2 - (y + 1)^2] - 10 (x/5 - x^3 - y^5) Exp[-x^2 - y^2] -
Exp[-(x + 1)^2 - y^2]/3 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9777138131620183,
"lm_q1q2_score": 0.8595539291032197,
"lm_q2_score": 0.8791467580102418,
"openwebmath_perplexity": 1176.8868397755964,
"openwebmath_score": 0.26597487926483154,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/113402/what-do-the-lines-in-a-contourplot-mean"
} |
With[{c = Subdivide[-5, 5, 10]},
GraphicsRow[{ContourPlot[peaks[x, y], {x, -3, 3}, {y, -3, 3},
ColorFunction -> "DarkTerrain",
Contours -> c, PlotRange -> All],
Plot3D[peaks[x, y], {x, -3, 3}, {y, -3, 3}, BoundaryStyle -> None,
Boxed -> False, ColorFunction -> "DarkTerrain",
MeshFunctions -> {#3 &}, Mesh -> {c}, PlotRange -> All,
ViewPoint -> {0, -2.7, 3.}]}]]
where I took the opportunity to use a function with more interesting contours. (Unfortunately, I don't know of any way to refine the meshlines produced by MeshFunctions; if you know a way, let me know in the comments!)
As noted in the comments above, the contours in ContourPlot[] are so-called isoclines; that is, lines of constant altitude ($z$ value). The interpretation of the colors depends on what colormap you're using; in this case, I used the "DarkTerrain" color gradient, which goes like this:
where the bluish color ("water") corresponds to the valleys, the whitish color ("snowcap") corresponds to the peaks, and the earthy colors corresponds to intermediate values. If you are familiar with map reading, running along isoclines corresponds to staying at a constant (flat) altitude, and crossing the isoclines corresponds to either ascent or descent.
A biographical segue: I was a Boy Scout once upon a time, and map reading was de rigueur for Scouts who wanted to have camping experience. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9777138131620183,
"lm_q1q2_score": 0.8595539291032197,
"lm_q2_score": 0.8791467580102418,
"openwebmath_perplexity": 1176.8868397755964,
"openwebmath_score": 0.26597487926483154,
"tags": null,
"url": "https://mathematica.stackexchange.com/questions/113402/what-do-the-lines-in-a-contourplot-mean"
} |
# Non-diagonal n-th roots of the identity matrix
Q. Are there $$n$$-th root analogs of this non-diagonal cube-root of the $$3 \times 3$$ identity matrix?
\begin{align*} \left( \begin{array}{ccc} 0 & 0 & -i \\ i & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right)^3 = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) \end{align*}
I am looking for $$A^n=I$$, where $$I$$ is any dimension $$\le n$$.
(A naive question: I am not an expert in this area.)
• I'm not sure what features you want of your "analogue" here. Are you just looking for a non-diagonal $A$ such that $A^n = I$? Does it have to be $3 \times 3$? Or maybe $n \times n$? Does it have to match some feature of how the non-zero terms are laid out (e.g. one non-zero term in each row/column)? It would be good to get some more information about what you'd like to see in an answer. Jul 1 at 20:26
• The problem I see is that a matrix can have n different nth roots of a matrix. How do you decide which you want? Jul 1 at 20:38
• @GeorgeIvey "The problem I see is that a matrix can have n different nth roots of a matrix." More, usually, and occasionally fewer. Jul 1 at 20:41
• The companion matrix of the $n$th cyclotomic polynomial has order $\phi(n)$, which is less than $n$. Does that work for you?
– lhf
Jul 1 at 20:48
• Hi, Joseph. The permutation matrix of a cyclic permutation always works. Apparently you can alter the final pair of 1's into $i$ and $-i$ and still get $A^n = I$ Jul 1 at 21:19
We want to solve $$A^n = I$$ where $$A$$ and $$I$$ are $$m \times m$$
Let $$A = P \Lambda P^{-1}$$
where $$\Lambda = \text{diag}(\omega_1, \omega_2, \dots , \omega_m)$$
where the $$\omega_k$$'s are $$n$$-roots of unity. That is, $$\omega_k = \exp\big(\dfrac{i 2 \pi j}{n} \big)$$ ,$$k = 1,2,\dots,m$$ and $$j \in \{ 0, 1, 2, ...., n-1 \}$$
and $$P$$ is any invertible $$m \times m$$ matrix. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534392852384,
"lm_q1q2_score": 0.8595441613979325,
"lm_q2_score": 0.8757869981319862,
"openwebmath_perplexity": 333.5576481456069,
"openwebmath_score": 0.9979684352874756,
"tags": null,
"url": "https://math.stackexchange.com/questions/4484433/non-diagonal-n-th-roots-of-the-identity-matrix/4484681#4484681"
} |
and $$P$$ is any invertible $$m \times m$$ matrix.
Then $$A^n = \big(P \Lambda P^{-1}\big)\big(P \Lambda P^{-1}\big)\dots \big(P \Lambda P^{-1}\big) = P^{-1} \Lambda^n P = P^{-1} I P = I$$
• Yours is the most general example, even in dimension $n$. Can you prove it?
– Ruy
Jul 2 at 0:26
• Thanks. Please check my updated solution for the proof. Jul 2 at 1:13
You can take a rotation matrix that rotates $$\phi=2\pi/n$$ around some axis, for example in 3 dimensions:
$$R_\phi=\begin{pmatrix} \cos \phi & \sin\phi & 0 \\ -\sin\phi & \cos\phi &0 \\ 0 & 0 & 1\\ \end{pmatrix}$$
This won't work for $$n=2$$ though, because $$R_\phi$$ is diagonal in that case like $$R_\phi=\operatorname{diag}(-1,-1,1)$$.
Then $$R_\phi^n=I$$ and $$R_\phi^k\neq I$$ for any $$0. You can build new matrices using any invertible matrix $$A$$ and conjugate like
$$R_{\phi,A}:= AR_\phi A^{-1} \tag 1$$
then obviously:
$$R_{\phi,A}^n = (AR_\phi A^{-1})^n = A R_\phi^nA^{-1} = I$$
(I am not sure whether this could fix the case $$n=2$$ and produce some valid (non-diagonal) results.)
• You can use Wolfram Alpha to check that {{1,1,0},{0,0,1},{1,0,0}}^-1 * {{-1,0,0},{0,-1,0},{0,0,1}} * {{1,1,0},{0,0,1},{1,0,0}} is not diagonal. Of course, for n=2 you can also use a reflection instead of a rotation, I.e. exchange two basis vectors. Jul 2 at 14:39
If $$A^n = I$$, i.e., $$A^n - I = 0$$, then the minimal polynomial $$m_A$$ of $$A$$ divides $$p(x) := x^n - 1$$. (For $$n > 1$$) one nondiagonal solution is the companion matrix of $$p$$ itself, namely, $$C_p := \pmatrix{\cdot & 1 \\ I_{n - 1} & \cdot}$$ (indeed, $$p = m_{C_p}$$).
Notice that $$C_p$$ is the permutation matrix for the permutation (in fact $$n$$-cycle) $$\pmatrix{1 & 2 & \cdots & n - 1 & n}$$ of $$n$$ elements. More generally, the permutation matrix of any permutation of $$n$$ elements of order dividing $$n$$ (and not the identity permutation) yields another solution. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534392852384,
"lm_q1q2_score": 0.8595441613979325,
"lm_q2_score": 0.8757869981319862,
"openwebmath_perplexity": 333.5576481456069,
"openwebmath_score": 0.9979684352874756,
"tags": null,
"url": "https://math.stackexchange.com/questions/4484433/non-diagonal-n-th-roots-of-the-identity-matrix/4484681#4484681"
} |
There are uncountably many solutions for any $$n > 1$$: Given any solution $$A$$ and any invertible matrix $$S$$, $$SAS^{-1}$$ is another solution provided it is not diagonal (since diagonality is a closed condition that by hypothesis is not always satisfied).
The most general solution of the equation $$A^n=1$$ among $$m\times m$$ complex matrices is $$A=SDS^{-1},$$ where $$S$$ is an invertible matrix and $$D$$ is a diagonal matrix whose diagonal entries are $$n^{th}$$ roots of unity.
The reason is that the minimal polynomial of such a matrix $$A$$ divides the polynomial $$x^n-1$$, and hence has distinct roots, which in turn implies that $$A$$ is diagonalizable. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534392852384,
"lm_q1q2_score": 0.8595441613979325,
"lm_q2_score": 0.8757869981319862,
"openwebmath_perplexity": 333.5576481456069,
"openwebmath_score": 0.9979684352874756,
"tags": null,
"url": "https://math.stackexchange.com/questions/4484433/non-diagonal-n-th-roots-of-the-identity-matrix/4484681#4484681"
} |
# Find a cubic polynomial.
If $f(x)$ is a polynomial of degree three with leading coefficient $1$ such that $f(1)=1$, $f(2)=4$, $f(3)=9$, then $f(4)=?,\ f(6/5)=(6/5)^3?$
I attempt:
I managed to solve this by assuming polynomial to be of the form $f(x)=x^3+ax^2+bx+c$, then getting the value of $a,b,c$, back substituting in the equation and so on....
But we can see:
$f(x)=q_1(x-1)+1\\f(x)=q_2(x-2)+4\\f(x)=q_3(x-3)+9$
Also when we put $x=1$ we get $f(1)=1^2$, when $x=2$ then $f(2)=2^2$, when $x=3$ then $f(3)=3^2$ but $f(4)\neq4^2$ (from answer).
Can this information be used to reproduce $f(x)$ directly without using the step I described in very first line of my solution?
Consider $g(x) = f(x)-x^2$. Then $g(1) = g(2) = g(3) = 0$ and $g$ is also a cubic polynomial and has leading coefficient 1. Thus $g(x) = (x-1)(x-2)(x-3)$ and hence $f(x) = (x-1)(x-2)(x-3)+x^2$. It now follows that $f(4) = 22$. Other values can be calculated.
• should $f(4)=22$ Jul 11 '17 at 3:58
• The OP wrote "with leading coefficient $1$". Jul 11 '17 at 3:59
• I assumed that $f(6/5)=(6/5)^3$ to be proved from the given information that $f$ is monic.
– user348749
Jul 11 '17 at 4:00
Classic long method:
Let $f(x) = x^3 + b x^2 + c x + d$ with $f(1) = 1$, $f(2) = 4$, $f(3) = 9$, which leads to \begin{align} f(1) &= 1 = 1 + b + c + d \hspace{10mm} \to d = -b - c \\ f(2) &= 4 = 8 + 4 b + 2 c + d = 8 + 3b + c \hspace{10mm} \to c = -4 - 3b, \, d = 4 + 2b \\ f(3) &= 9 = 27 + 9b + 3c + d = 19 + 2b \end{align} from which $b = -5$, $c = 11$, and $d = -6$ and $$f(x) = x^3 - 5 \, x^2 + 11 \, x -6.$$
With $f(x)$ then \begin{align} f(4) &= 64 - 80 + 55 -6 = 22 \\ f\left(\frac{6}{5}\right) &= \left(\frac{6}{5}\right)^{3} - \frac{36 - 66 + 30}{5} = \left(\frac{6}{5}\right)^{3}. \end{align} | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534333179648,
"lm_q1q2_score": 0.859544156171872,
"lm_q2_score": 0.8757869981319863,
"openwebmath_perplexity": 268.9100203312175,
"openwebmath_score": 1.0000100135803223,
"tags": null,
"url": "https://math.stackexchange.com/questions/2354401/find-a-cubic-polynomial"
} |
It may also be noticed that $f(x)$ can be seen in the form $$f(x) = \left(x - \frac{5}{3}\right)^{3} + \frac{8}{3} \, \left( x - \frac{5}{3}\right) + \frac{83}{27}.$$ From this it is easy to see that \begin{align} f\left(\frac{5}{3}\right) &= 3 + \frac{2}{27} \\ f\left(\frac{5}{6}\right) &= \frac{59}{216}. \end{align}
• Looking at your last form of $f(x)$ if I put $5/3$ in then I would not get the same answer as you have for $f(5/3)$ just after - the former gives $3+2/27$... I've not chased the maths to see which is right, just happened to notice that discrepancy (or I've done something stupid which is also possible). Jul 11 '17 at 8:47
If
• $$f(x) = x^3 + ax^2 + bx + c$$
• $$f(1) = 1$$
• $$f(2) = 4$$
• $$f(3) = 9$$
then
\begin{align} f(x+1) - f(x) &= (3x^2+3x+1) + a(2x+1) + b \\ \hline 4-1 &= f(2)-f(1) \\ 3 &= 7 + 3a + b \\ \hline 9-4 &= f(3) - f(2) \\ 5 &= 19 + 5a + b \\ \hline 3a + b &= -4 \\ 5a + b &= -14 \\ \hline a &= -5 \\ b &= 11 \\ c &= -6 \end{align}
S0 $$f(x) = x^3 - 5x^2 + 11x - 6$$.
You could make a difference table
$$\begin{array}{c} 1 && 4 && 9 \\ & 3 && 5 \\ && 2 \end{array}$$
Using $$f(x) = x^3 + ax^2 + bx + c$$, this corresponds to
$$\begin{array}{c} 1+a+b+c && 8 + 4a + 2b + c && 27 + 9a + 3b + c \\ & 7+3a+b && 19 + 5a + b \\ && 12+2a \end{array}$$
Then, comparing entries...
\begin{align} 12+2a=2 &\implies a = -5 \\ 7+3a+b = 3 &\implies b=11 \\ 1+a+b+c = 1 &\implies c = -6 \end{align} | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534333179648,
"lm_q1q2_score": 0.859544156171872,
"lm_q2_score": 0.8757869981319863,
"openwebmath_perplexity": 268.9100203312175,
"openwebmath_score": 1.0000100135803223,
"tags": null,
"url": "https://math.stackexchange.com/questions/2354401/find-a-cubic-polynomial"
} |
# Analyzing $\biggl\lfloor{\frac{x}{5}}\bigg\rfloor=\bigg\lfloor{\frac{x}{7}}\bigg\rfloor$
How many non negative integral values of $x$ satisfy the equation :$$\biggl\lfloor{\dfrac{x}{5}}\bigg\rfloor=\bigg\lfloor{\dfrac{x}{7}}\bigg\rfloor$$.
My try:
Writing few numbers and putting in the required equation.
$\underbrace{0,1,2,3,4}_{\implies 0=0},\underbrace{5,6}_{\displaystyle\times},\underbrace{7,8,9}_{\implies 1=1},\underbrace{10,11,12,13}_{\displaystyle\times} ,\underbrace{14}_{\implies 2=2},\underbrace{15,16,17,18,19,20}_{\text{LHS gives 3 but RHS gives 2}},\underbrace{21,22,23,24}_{\text{LHS gives 4 but RHS gives 3}}$.
The point which I wish to make here is that after $14$ we'll get $5's$ multiple before than $7's$, hence $\forall x \geq 15 \implies \biggl\lfloor{\dfrac{x}{5}}\bigg\rfloor>\bigg\lfloor{\dfrac{x}{7}}\bigg\rfloor$
Hence the only solution are: $\{0,1,2,3,4,7,8,9,14\}$. Does these make sense?
I need to know other solutions that don't incorporate such analyzing, means using some properties and framing it to solve above.
By the division rule,
$$x=5q+r=7q+s$$ with $0\le r<5,0\le s<7$.
From this we draw $$2q=r-s\in[-6,4],$$
thus the only possibilities are with $q=0,1,2$.
• $q=0$: $r=s$, possible for $x\in[0,4]\cap[0,6]=[0,4]$;
• $q=1$: $5+r=7+s$, possible for $x\in[5,9]\cap[7,13]=[7,9]$;
• $q=2$: $10+r=14+s$, possible for $x\in[10,14]\cap[14,20]=[14]$.
• why intersection? – mnulb Jan 25 '17 at 16:52
• Because the two bracketings must hold. What else would you do ? – Yves Daoust Jan 25 '17 at 16:53
• did you came up with those intersections by hit and trials or something different, and if something new, please don't mind adding there. – mnulb Jan 25 '17 at 17:08
• @Ayushakj: the intervals are drawn from the equations and inequations. – Yves Daoust Jan 25 '17 at 17:11 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534338604443,
"lm_q1q2_score": 0.8595441518741312,
"lm_q2_score": 0.8757869932689566,
"openwebmath_perplexity": 887.7086915685115,
"openwebmath_score": 0.9556778073310852,
"tags": null,
"url": "https://math.stackexchange.com/questions/2113526/analyzing-biggl-lfloor-fracx5-bigg-rfloor-bigg-lfloor-fracx7-bigg"
} |
Rewrite the given condition as $$5n\le x<5n+5,\quad 7n\le x<7n+7$$ where $n$ is an integer. Since $x\ge0$, $n\ge0$ as well, so the above is equivalent to $7n\le x<5n+5$. The only nonnegative integers $n$ satisfying $7n<5n+5$ are $0$, $1$, and $2$. And so just a little cleanup is needed.
For a solution the LHS and RHS are a positive integer, let's call this $c$: $$c = \left \lfloor \frac{x}{5} \right \rfloor = \left \lfloor \frac{x}{7} \right \rfloor$$
Then, there are integers $0 \leq a < 5, 0 \leq b < 7$, s.t.: $$x = 5c + a = 7c + b$$ Thus: $$a - b = 2c$$
With the restrictions to $a,b$, this is enough to find all solutions.
Analysis:
Let us take the general form where $p>q$ :
$$\left\lfloor{\dfrac{x}{q}}\right\rfloor=\left\lfloor{\dfrac{x}{p}}\right\rfloor$$
Solution for above equation exists when $\dfrac{x}{q}-\dfrac{x}{p}<1\Rightarrow x<\dfrac{p\cdot q}{p-q}$
So there is no need to check for integers greater than$\left\lfloor\dfrac{p\cdot q}{p-q}\right\rfloor$
Also $\left\lfloor{\dfrac{x}{q}}\right\rfloor=\left\lfloor{\dfrac{x}{p}}\right\rfloor=k$ implies \begin{align} &x\in [k\cdot p,(k+1)p) \cap\ [k\cdot q,(k+1)\cdot q)\\ \Rightarrow\ \ &x\in [k\cdot p,(k+1)\cdot q) \end{align}
The integers which satisfy the equation are in $$[0p,1q)\ \cup\ [1p,2q)\ \cup\ [2p,3q) \cup\ ...\ \cup\ [(n-1)p,nq)$$ where $nq<\left\lfloor\dfrac{p\cdot q}{p-q}\right\rfloor$ .
• Well done but you are missing the case $k=0$, giving the interval $[0,q)$. – Yves Daoust Jan 26 '17 at 7:41 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9814534338604443,
"lm_q1q2_score": 0.8595441518741312,
"lm_q2_score": 0.8757869932689566,
"openwebmath_perplexity": 887.7086915685115,
"openwebmath_score": 0.9556778073310852,
"tags": null,
"url": "https://math.stackexchange.com/questions/2113526/analyzing-biggl-lfloor-fracx5-bigg-rfloor-bigg-lfloor-fracx7-bigg"
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.