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No, you need to divide by $\displaystyle 12^3$...essentially a 12 for each of the 3 spatial dimensions.
$\displaystyle 7128\text{ in}^3\cdot\left(\frac{1\text{ ft}}{12\text{ in}} \right)^3=\frac{7128}{12^3}\,\text{ft}^3=156.75 \text{ ft}^3$
Oh, ok that makes since. So for part 5 would I use the formula 1/2*b*h so 1/2*(99)*(104) = 5148 in^2
No, you have a trapezoid, not a triangle. You want to use:
$\displaystyle A=\frac{h}{2}(B+b)$
where:
$\displaystyle h=99$
$\displaystyle B=104$
$\displaystyle b=32$
O.k., sorry for not getting this right away, thanks for your patience lol.
So the answer would then be 6732
An impatient person has no business trying to help on a forum. You are doing fine.
Yes, that is the correct answer. Now, in order to find this area using integration, I recommend orienting your coordinate axes such that the origin is at the bottom corner under the lowest point of the ceiling. We will then want to write the upper slanting edge of the wall as a linear function. What would the y-intercept and the slope be?
Would the slope be 8 and the y-intercept 32?
You have the correct intercept, but for the slope, think of the rise over run of an individual step.
ok so the slope would be 8/11?
Yes, good work! So, what would the linear function representing the slanted edge be?
(8/11)x+32?
Exactly! Now, over what interval do you want to integrate?
Would it be from 5 to 13?
and if that is the interval would it be:
(4/11)x^2+32x evaluated from 5 to 13= 308.364 | {
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1. ## Modulo problem
Okay, so I'm supposed to find the least nonnegative integer i such that
4^30 is congruent to i (mod 19).
I'm not sure how to approach this since my textbook doesn't really say how this type of problem is done so any help would be greatly appreciated.
2. Originally Posted by clockingly
Okay, so I'm supposed to find the least nonnegative integer i such that
4^30 is congruent to i (mod 19).
I'm not sure how to approach this since my textbook doesn't really say how this type of problem is done so any help would be greatly appreciated.
My approach here won't be unique, but the method should be fairly clear.
$4^3 = 64 \equiv 7$ (mod 19)
So
$4^{30} = (4^3)^{10} \equiv 7^{10}$ (mod 19)
Again:
$7^2 \equiv 11$ (mod 19)
So
$7^{10} = (7^2)^{5} \equiv 11^5$ (mod 19)
You can probably do this one by directly but we can simplify this one last time:
$11^2 \equiv 7$ (mod 19)
$11^3 \equiv 1$ (mod 19)
So finally:
$4^{30} \equiv 7^{10} \equiv 11^5 = 11^2 \cdot 11^3 \equiv 7 \cdot 1 = 7$ (mod 19)
$4^{30} \equiv 7$ (mod 19)
-Dan
I am still a little confused as to how you got the very first equation though (listd below)...I'm a little confused as to where the 7 came from...
4^3 = 64 is congruent to 7(mod 19)
4. Originally Posted by clockingly
Okay, so I'm supposed to find the least nonnegative integer i such that
4^30 is congruent to i (mod 19).
I'm not sure how to approach this since my textbook doesn't really say how this type of problem is done so any help would be greatly appreciated.
You can also do this...
$4^{30}=2^{60}$
And,
$2^{18}\equiv 1 (\mbox{ mod }19)$
By Fermat's little theorem since 19 is prime not divisible by 2.
Cubing,
$2^{54}\equiv 1 (\mbox{ mod }19)$
Multiply both sides by $2^6=64\equiv 7$ we we have,
$2^{60}\equiv 7 (\mbox{ mod }19)$
5. Originally Posted by clockingly
I am still a little confused as to how you got the very first equation though (listd below)...I'm a little confused as to where the 7 came from... | {
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4^3 = 64 is congruent to 7(mod 19)
That is becuase,
$64\equiv 7 (\mbox{ mod }19)$
Right?
Because it leaves a remainder of 19.
( $64=3\cdot 19+7$)
6. Hello, clockingly!
Another approach . . .
Find the least nonnegative integer $n$ such that: . $4^{30} \equiv n \pmod{19}$
I found that: . $4^5 \:=\:1024 \:\equiv\:-2 \pmod{19}$
Then: . $4^{30} \:=\:(4^5)^6 \:\equiv\:(-2)^6 \pmod{19}$
And: . $(-2)^6\:=\:64\:\equiv\:7\pmod{19}$
Therefore: . $4^{30}\:\equiv\:7\pmod{19}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I also found that: . $4^9\:\equiv\:1\pmod{19}$
Then: . $4^{30}\:=\:\left(4^9\right)^3\cdot4^3 \:\equiv\:(1^3)4^3\pmod{19}$
And: . $4^3\:=\:64\:\equiv\:7\pmod{19}$
But I see that TPHacker beat me to it. | {
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# What is the least amount of pieces on a board with the following conditions:
There's an infinite board. Imagine you add a rectangle of $m*n$ pieces. With $m,n \geq 2$ (There's a piece every square, and you can't put one above other.) You can make a piece 'jump' other that is next to it (Vertically or horizontally only, not diagonally), if the square next to the "jumped" piece (In the same direction) is empty, also, the "jumped" piece disappears. After many movements, What is the least amount of pieces that can be on the board?
So, it's obvious that the answer can't be $0$ since there's no point of it, it has to be at least one if we have, for example, to pieces left of all the $mn$ pieces that are connected each other. I tried to do a coloring of the infinite board, in a chess pattern, so we have Black pieces (Pieces in a black tile) and White pieces, this make any white piece unable to make dissapear another white one and the same with black ones. At the end, the least we have of each color the better, because if we get to have only two left with different colors and connected, then we'll know the least number of pieces is 1. i tried looking for different rectangles of pieces, and i got 1 and 2 as answers. For example doing the following when $m=3$ and $n=4$ makes the board have 1 piece at the end: (From left to right)
It clearly follows that it has 1 piece at the end. I found other cases as ($m=2,n=4$), ($m=2, n=5$), but i also found cases where i got the least is 2, like ($m=2, n=3$), ($m=3, n=3$), etc. i haven't noticed anything else besides this. Any suggestions?
• How did you get from your second board to your third? – Mike Earnest Aug 21 '18 at 17:46
• Move (3,1) to (5,1) an then move (1,1) to (3,1) – SonodaUmi Aug 21 '18 at 17:59
• Then you have a mistake; the checker at (4,1) was hopped over, so should be removed. – Mike Earnest Aug 21 '18 at 18:01
This is a fun problem with a nice solution! | {
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This is a fun problem with a nice solution!
First of all, imagine coloring the checkerboard in three colors like so:
A B C A B C A B C ...
B C A B C A B C A ...
C A B C A B C A B ...
...
Let $a,b,c$ be the number of checkers on each color. Each move increases one of these variables by $1$, and decreases the others by $1$. This means that the quantities $a-b,b-c$ and $c-a$ will always have the same parity.
Now, suppose that either $m$ or $n$ is a multiple of $3$. Then you will initially have $a=b=c$, so that $a-b,b-c$ and $c-a$ all start out even. However, if you manage to reduce the board to a single checker, then at least one of $a-b,b-c$ or $c-a$ would be odd. Therefore, the problem is insoluble when $m$ or $n$ is a multiple of $3$.
When neither dimension is a multiple of $3$, you can succeed. Consider the following "T"-move. It allows you to eliminate a $3\times 1$ block of marbles, provided one of the ends has a marble on on side and an empty space on the other:
• • _ _ _ • _ • • • _ _
• --> • --> _ --> _
• • _ _
Applying this repeatedly, you eliminate a $3\times n$ block from the top of the rectangle, provided $m>3$ and $n\ge 3$:
• • • • • • • •
• • • • • • • •
• • • • • • • •
• • • • • • • •
|
v
• • • • • • • _
• • • • • • • _
• • • • • • • _
• • • • • • • •
|
v
• • • • • • _ _
• • • • • • _ _
• • • • • • _ _
• • • • • • • •
|
v
...
|
v
• • • _ _ _ _ _
• • • _ _ _ _ _
• • • _ _ _ _ _
• • • • • • • •
and then apply the same idea with three sideways T's to get rid of that last $3\times 3$ block. The same idea allows you eliminate three columns at a time provided the dimensions are larger enough.
This procedure allows you to reduce the board either a $4\times 4$, $2\times 4$, $4\times 2$ or $5\times 5$ rectangle, depending on the remainders of $m$ and $n$ modulo $3$. I leave it to you to figure out how to solve these small cases. | {
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• My teacher said it was a really beautiful solution, and it really is. Thank you! – SonodaUmi Aug 21 '18 at 18:25
• @cptnSonoda Does your teacher sit next to you and checked the answer on math.stackexchange (MSE)? – callculus Aug 21 '18 at 18:38 | {
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# What is a good upper bound $n^n(n-1)^{n-1}\ldots2^21^1$?
Given an integer $n \ge 1$, I'd like to have a not-very-loose upper bound for the integer $$u(n) := \Pi_{k=1}^n k^k = n^n(n-1)^{(n-1)}\ldots2^21^1.$$
It's easy see that, $u(n) \le n^{n(n+1)/2}$, but this is not very interesting.
# Update
We have $u(n) \le e^{\left(\frac{1}{2}n(n+1)\log\left(\frac{2n + 1}{3}\right)\right)}$, and we can't really do much better!
Indeed, using Euler-Maclaurin, we have
$\log(u(n)) = \int_2^nx\log x dx = \frac{1}{4}n^2(2\log(n) - 1) - 2\log(2) + \frac{1}{4} + \text{error terms}$, which is comparable to the bound $\log(u(n)) \le \frac{1}{2}n(n+1)\log\left(\frac{2n + 1}{3}\right)$ in the accepted answer (see below). In particular, we can conclude that accepted answer's bound is tight!
• Taking logarithms may be a good start whenever you're dealing with products. Sep 9, 2016 at 13:45
• Ya, I contemplated the sequence $\sum_{k=1}^nk \log(k)$, but nothing pops-up... Sep 9, 2016 at 13:46
• Yes. Take logarithms, use Euler-MacLaurin, and exponentiate. Sep 9, 2016 at 13:47
• Thanks for pointing to Euler-Maclaurin. At the moment, I was just thinking about the Abel summation formula, but your suggestion looks more appropriate. Sep 9, 2016 at 13:52
• See also $(7)$ here: mathworld.wolfram.com/Hyperfactorial.html Sep 9, 2016 at 17:24
Using Jensen's inequality:
Letting $A= \sum_{k=1}^n k = \frac{n (n+1)}{2}$ and $B= \sum_{k=1}^n k^2 = \frac{n (n+1)(2n+1)}{6}$
We have
\begin{align} \log(u(n)) &=\sum_{k=1}^n k \log(k) \\ &= A \sum_{k=1}^n \frac{k}{A} \log(k) \tag{1}\\ &\le A \log \left( \sum_{k=1}^n \frac{k}{A} k \right) = A \log \left( \frac{B}{A} \right) \tag{2} \end{align}
Hence
$$\log(u(n)) \le \frac{n(n+1)}{2} \log\left(\frac{2 n+1}{3}\right) \tag{3}$$
The bound seems to be quite tight:
Update: as noted by comments and OP, the bound $(3)$ agrees with the true order of growth; this can be checked by applying the trapezoidal rule to the integral: | {
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$$-\frac{1}{4}=\int_{0}^{1} x \log(x) dx \approx \frac{1}{n+1} \sum_{k=1}^n \frac{k}{n} \log\left(\frac{k}{n}\right)$$ which gives
$$\log(u(n)) \approx\frac{ n(n+1)}{2}\left( \log n -\frac{1}{2} \right) \tag{4}$$
If one is interested in an approximation (instead of a bound), $(4)$ might be preferable.
• Perfect! Thanks. I see you've used the concave function $\phi(x) := log(x)$ and positive scalars $a_k = k$ in Jensen's inequality: en.wikipedia.org/wiki/Jensen%27s_inequality Sep 9, 2016 at 14:05
• Now that's a freaky tight bound! I Indeed, it's no surprise that the bound is tight. Using Euler-Maclaurin suggested by Daniel above, I can get $\log(u(n)) \approx \frac{1}{4} (n^2(2\log(n) - 1) - 1)$ :). Thus I don't need to beg the error terms in my EM, and can just use your bound instead. Thanks again. Sep 9, 2016 at 14:10
Another simple way for finding an upper bound is using the Abel's summation $$S=\sum_{k=1}^{n}k\log\left(k\right)=\frac{n\left(n+1\right)}{2}\log\left(n\right)-\frac{1}{2}\int_{1}^{n}\frac{\left\lfloor t\right\rfloor \left(\left\lfloor t\right\rfloor +1\right)}{t}dt$$ where $\left\lfloor t\right\rfloor$ is the floor function and since $\left\lfloor t\right\rfloor >t-1$ we get $$S<\frac{n\left(n+1\right)}{2}\log\left(n\right)-\frac{1}{2}\int_{1}^{n}\left(t-1\right)dt$$ $$=\color{red}{\frac{n\left(n+1\right)}{2}\log\left(n\right)-\frac{\left(n-1\right)^{2}}{4}}.$$
Don't have the reputation, or else this would be a comment. As mentioned in a link posted by leonbloy, the hyperfactorial, which shows up in the theory of the Barne's G function, is related to $\int_0^n \log\Gamma(x) dx.$ Good bounds on this should give a better bound than that found by Jensen's inequality. I found the expression
$$\log(u(n)) \le A(n):=\dfrac{(n+1/2)^2}{2}(\log(n+1/2)-3/2)+\dfrac{n(n+1)}{2}+ \dfrac{9}{8}\log(2/3)+\dfrac{11}{16}.$$
For a comparison, let's define 'Jensen' and 'A' ratios | {
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For a comparison, let's define 'Jensen' and 'A' ratios
$$R_J(n)=\dfrac{\log(u(n))}{n(n+1)/2\log((2n+1)/3))},\quad R_A(n)=\dfrac{\log(u(n))}{A(n)} .$$
Then (approximately) $R_J(100)=0.892$ , $R_A(100)=0.999992;$ and $R_J(10^5)=0.9915$ , $R_A(10^5)=0.99999999915$ (nine nines).
• Interesting. Do include some detail on your derivation of $A(n)$ (not just an expression for it...). Sep 13, 2016 at 7:39
• Great answer! Upvoted. Jan 23, 2017 at 7:22 | {
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# Factorise $x^5+x+1$
Factorise $$x^5+x+1$$
I'm being taught that one method to factorise this expression which is $=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1$
$=x^3(x^2+x+1)-x^2(x^2+x+1)+x^2+x+1$
=$(x^3-x^2+1)(x^2+x+1)$
My question:
Is there another method to factorise this as this solution it seems impossible to invent it?
• In general, factorization of polynomials is, as you say, "impossible to invent" by hand. There are algorithms, though, which can solve the problem. May 23, 2017 at 15:04
• Any high school student can solve this problem when he guesses one of the factors and then calculates the other factor by division. But that is not really a useful method to factor a polynomial. But it is similar to the method most of the answers uses. May 23, 2017 at 16:03
• $$5 \equiv 2 \pmod 3$$ May 23, 2017 at 16:06
• I just don't know how to accept only one answer when all the answers are simply amazing and awesome. May 23, 2017 at 22:25
• @Mathxx, See math.stackexchange.com/questions/1584594/… May 24, 2017 at 5:36
With algebraic identities, this is actually rather natural:
Remember if we had all powers of $x$ down to $x^0=1$, we could easily factor: $$x^5+x^4+x^3+x^2+x+1=\frac{x^6-1}{x-1}=\frac{(x^3-1)(x^3+1)}{x-1}=(x^2+x+1)(x^3+1),$$ so we can write: \begin{align}x^5+x+1&=(x^2+x+1)(x^3+1)-(x^4+x^3+x^2)\\ &=(x^2+x+1)(x^3+1)-x^2(x^2+x+1)\\ &=(x^2+x+1)(x^3+1-x^2). \end{align} | {
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• That's true, but it's not a factorisation. What do you mean? May 23, 2017 at 15:06
• What does "this is rather natural" mean in this context? May 23, 2017 at 15:42
• I simply meant any high school student which is trained to tackle formulae using various identities can factor this polynomial. May 23, 2017 at 15:45
• I think this is no better than the solution the user already showed us. One has to guess the right things. May 23, 2017 at 16:31
• Well, if you do not train, you'll never get a result. All I want to say is that everything is at the high school level, there's no hammersledge theorem, and that training in various ways stimulates imagination, which is the first quality to find proofs. May 23, 2017 at 16:35
If you suspect there exists a factorization over $\mathbb{Q}[x]$ into polynomials of degree 2 and 3, but you just don't know their coefficients, one way is to write it down as
$x^5 + x + 1 = (x^2 + ax + b)(x^3 + cx^2 + dx + e)$ where the coefficients are integers (by Gauss' lemma). And then expand and solve the system.
Then $a + c = 0, b + ac + d = 0, bc + ad + e = 0, ae + bd = 1, be = 1$. So we get $c = -a$ and $b = e = 1$ or $b = e = -1$.
In the first case we reduce to $1 - a^2 + d = 0, -a + ad + 1 = 0, a + d = 1$ which gives $d = 1 - a, 1 - a^2 + 1 - a = 0, 1 - a^2 = 0$ so $a = 1, b = 1, c = -1, d = 0, e = 1$.
Substituting gives us $x^5 + x + 1 = (x^2 + x + 1)(x^3 - x^2 + 1)$
If the factorization was not over $\mathbb{Q}[x]$ then things would get more complicated because I could not assume $b = e = +/- 1$.
• If you suspect there exists a factorization over Q[x] than it is easy to show that polynomial cannot be split into a polynomail of degree 4 and a linears factor, so it must be the product of thwo polynomails of degree 2 and 3 May 23, 2017 at 15:46 | {
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Alternatively note that \begin{align}x^5 + x + 1 &= x^5 - x^2 + x^2 + x + 1\\ & =x^2(x^3-1) + \color{red}{x^2+x+1} \\ & = x^2(x-1)\color{red}{(x^2+x+1)} + \color{red}{x^2+x+1} \\& =\color{blue}{(x^3-x^2+1)}\color{red}{(x^2+x+1)} \end{align}
where we used the well known identity $x^3 - 1 = (x-1)(x^2+x+1)$ in the third equality.
Meta: whilst the first step may seem arbitrary and magical, it is natural to want to insert a term like $x^2$ or $x^3$ into the equation in order to get some traction with factorising $x^5 + x^k$.
Note that if $z^3=1, z\neq 1$ so that $z^3-1=(z-1)(z^2+z+1)=0$ then $z^5+z+1=z^2+z+1=0$. A key to this observation is just seeing whether an appropriate root of unity may be a root.
This tells you that $x^2+x+1$ is a factor of $x^5+x+1$, and the question then is how you do the division. The factorisation method you have been shown is equivalent to doing the polynomial long division.
Another method, not suggested by others, is to use the fact that you know $x^2+x+1$ is a factor and multiply through by $x-1$ so that this factor becomes $x^3-1$.
So $(x-1)(x^5+x+1)=x^6-x^5+x^2-1=(x^3-1)(x^3+1)-x^2(x^3-1)=(x^3-1)(x^3-x^2+1)=(x-1)(x^2+x+1)(x^3-x^1+1)$
• I think this is no better than the solution the user already showed us. One has to guess the right things May 23, 2017 at 16:32
• @miracle173 But the user suggested the solution could not be discovered easily. The art of problem solving is in part learning the kind of things you might try when faced with something unfamiliar. May 23, 2017 at 19:22
standard trick from contests: $5 \equiv 2 \pmod 3.$ Therefore, if $\omega^3 = 1$ but $\omega \neq 1,$ we get $$\omega^5 = \omega^2$$ $$\omega^5 + \omega + 1 = \omega^2 + \omega + 1 = 0$$ Which means, various ways of saying this, $x^5 + x + 1$ must be divisible by $$(x - \omega)(x - \omega^2) = x^2 + x + 1.$$ This is what we call a "minimal polynomial" for $\omega$ | {
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The same idea would work for $$x^{509} + x^{73} + 1$$ although the other factor would be worse
SEE
Prime factor of $A=14^7+14^2+1$
• I think this is no better than the solution the user already showed us. One has to guess the right things May 23, 2017 at 16:32
• @miracle173 it's a matter of where the questions arise. This is a contest or contest preparation question, there is a trick; this comes up over and over again on this site, from kids who are trying to get better at contests. You might ask the OP where he got the question and what is the setting responsible for "I'm being taught..." Put another way, someone skilled at searching could find hundreds of questions on MSE where one factor is $x^2 + x + 1$ May 23, 2017 at 16:42
So far the answers just (cleverly) elaborate on high school tricks and techniques. Therefore, I think it can be interesting to see, instead, how standard modern algorithms work in this special case. I will implement a small version of the Berlekamp-Zassenhaus algorithm. I will try to factor $F(x)=x^5+x+1$ over $\mathbb Z[x]$ as a product $f_1(x)f_2(x)$ of polynomials of degree 2 and 3; it will not be long (of course), nor difficult. I recall what is the plan:
• Bound the coefficients of the factors $f_1,f_2$;
• Factor $F(x)=g_1(x)g_2(x)$ over modulo $p$ for some prime $p$;
• Lift (in essence, by Hensel's lemma) the factorization modulo higher $p^k$.
Bound the coefficients of $f_1(x)$: The leading coefficient of $F$ is $c=1$, the degree of $f_1$ is $\delta=2$, and the roots $\alpha$ of $F$ satisfy ${|\alpha|}^5\leq 1+|\alpha|$, so for sure, say, $|\alpha|<\rho=1.5$. Therefore the (absolute values of the) coefficients of $f_1(x)$ are all dominated by those of $(x+\rho)^2$. In particular they are all $<3$. This is called the binomial bound. The same estimate can be obtained with the Knuth-Cohen bound. See Abbott, John. "Bounds on Factors in Z [x]." Journal of Symbolic Computation 50 (2013): 532-563. | {
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Find $g_i(x)\equiv f_i(x)$ modulo 2: Since $F(0)=1$ and $F(1)=3$ we have that $g_1(0)=g_1(1)=1\bmod 2$. Thus the only possibility is $g_1(x)=x^2+x+1$. By polynomial long division in $\mathbb F_2[x]$ we get $g_2(x)=\frac{F(x)}{g_1(x)}=x^3+x^2+1$. Well, if you are clever enough, and not a computer, you might finish the exercise here, by doing long division in $\mathbb Z[x]$.
Factor modulo 4 as $F=(g_1+2 h_1)(g_2 + 2 h_2)$: In other words, we need $g_1 h_2+g_2 h_1 = \frac {F(x)-g_1(x)g_2(x)}{2} = x^4+x^3+x^2 \bmod 2$. Of course the solution is $h_1=0$ and $h_2=x^2$.
Conclusion: We have $f_1(x)\equiv x^2+x+1\bmod 4$. Since the absolute value of the coefficients of $f_1(x)$ is at most $2$, we get $f_1(x)= x^2+x+1$. It works.
Supplement: actually I am deeply convinced that the most natural technique to factor $F(x)$ is the one provided by the OP (although all the other approaches, included the one I described above, are interesting). I'll try to justify this claim. Suppose you want to factor $7763073514021$ in prime numbers. The factor $7$ is easy to find (actually, this completes the factorization). Why? Because you decompose $7-7-63-0-7-35-14-0-21$ and you factor out termwise. "Piecewise" factorization is by far the most natural, and easy to spot, approach to factorization "by hand". Another example is $x^6-x^4-20x^3+14x^2+20 x -14$, where you may wish to take advantage of the pattern $(1,-1),(-20,20),(14,-14)$. Now, suppose you want to factor the number $636362236363$. It's very similar to the example before, only that you must be able to see the "negative": you are just subtracting a "14" from $636363636363$. Although the pattern of coefficients $[1,0,0,0,1,1]$ of $F(x)$ might be irregular at first glance, I find it very natural to see it as a $0-111-00$ subtracted from a $111-111$. | {
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• Or else (but only for "how to factor", not "why does it factor"): coefficients of f_1 have norm at most 2; F is monic; F(0)=F(-1)=1. So f_1=x^2+x+1 or f_1=x^2-x-1. Now use F(-2)=-33. May 24, 2017 at 1:09
Let $f (x)=x^5+x+1$
$$f'(x)=5x^4+1>0$$
$f$ is stricly increasing at $\mathbb R$, thus it has only one real root $\alpha \in (-1,0)$.
the factorisation will be of the form
$$(x-\alpha)(x^2+ax+b)(x^2+cx+d)$$
with $a^2-4b <0$ and $c^2-4d <0$.
• Why can't it be factored in $\mathbf R[X]$? The only irreducible polynomials in $\mathbf R[X]$ are linear polynomials and quadratic polynomials with a negative discriminant. May 23, 2017 at 14:56
• Just because it has no linear factors does not mean that it does not have higher order factors (DAMHIKT). Also, he showed a factorization of it and the factorization is correct: just multiply it out. So, your answer is wrong: it can be factorized in $\mathbb R[X]$. May 23, 2017 at 14:58
• @Nick Yes you are right. i edited. May 23, 2017 at 15:02 | {
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Binary code with constraint
Suppose I have an alphabet of n symbols. I can efficiently encode them with $\lceil \log_2n\rceil$-bits strings. For instance if n=8:
A: 0 0 0
B: 0 0 1
C: 0 1 0
D: 0 1 1
E: 1 0 0
F: 1 0 1
G: 1 1 0
H: 1 1 1
Now I have the additional constraint that each column must contain at most p bits set to 1. For instance for p=2 (and n=8), a possible solution is:
A: 0 0 0 0 0
B: 0 0 0 0 1
C: 0 0 1 0 0
D: 0 0 1 1 0
E: 0 1 0 0 0
F: 0 1 0 1 0
G: 1 0 0 0 0
H: 1 0 0 0 1
Given n and p, does an algorithm exist to find an optimal encoding (shortest length) ? (and can it be proved that it computes an optimal solution?)
EDIT
Two approaches have been proposed so far to estimate a lower bound on the number of bits $m$. The goal of this section is to provide an analysis and a comparaison of the two answers, in order to explain the choice for the best answer.
Yuval's approach is based on entropy and provides a very nice lower bound: $\frac{logn}{h(p/n)}$ where $h(x) = xlogx + (1-x)log(x)$.
Alex's approach is based on combinatorics. If we develop his reasonning a bit more, it is also possible to compute a very good lower bound:
Given $m$ the number of bits $\geq\lceil log_2(n)\rceil$, there exists a unique $k$ such that $$1+\binom{m}{1} + ... +\binom{m}{k} \lt n \leq 1+\binom{m}{1} + ... + \binom{m}{k}+\binom{m}{k+1}$$ One can convince himself that an optimal solution will use the codeword with all bits low, then the codewords with 1 bit high, 2 bits high, ..., k bits high. For the $n-1-\binom{m}{1}-...-\binom{m}{k}$ remaining symbols to encode, it is not clear at all which codewords it is optimal to use but, for sure the weights $w_i$ of each column will be bigger than they would be if we could use only codewords with $k+1$ bits high and have $|w_i - w_j| \leq 1$ for all $i, j$. Therefore one can lower bound $p=max(w_i)$ with $$p_m = 0 + 1 + \binom{m-1}{2} +... + \binom{m-1}{k-1} + \lceil \frac{(n-1-\binom{m}{1}-...-\binom{m}{k}) (k+1)}{m} \rceil$$ | {
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Now, given $n$ and $p$, try to estimate $m$. We know that $p_m \leq p$ so if $p \lt p_{m'}$, then $m' \lt m$. This gives the lower bound for $m$. First compute the $p_m$ then find the biggest $m'$ such that $p \lt p_{m'}$
This is what we obtain if we plot, for $n=1000$, the two lower bounds together, the lower bound based on entropy in green, the one based on the combinatorics reasonning above in blue, we get:
Both look very similar. However if we plot the difference between the two lower bounds, it is clear that the lower bound based on combinatorics reasonning is better overall, especially for small values of $p$.
I believe that the problem comes from the fact that the inequality $H(X) \leq \sum H(X_i)$ is weaker when $p$ gets smaller, because the individual coordinates become correlated with small $p$. However this is still a very good lower bound when $p=\Omega(n)$.
Here is the script (python3) that was used to compute the lower bounds:
from scipy.misc import comb
from math import log, ceil, floor
from matplotlib.pyplot import plot, show, legend, xlabel, ylabel
# compute p_m
def lowerp(n, m):
acc = 1
k = 0
while acc + comb(m, k+1) < n:
acc+=comb(m, k+1)
k+=1
pm = 0
for i in range(k):
pm += comb(m-1, i)
return pm + ceil((n-acc)*(k+1)/m)
if __name__ == '__main__':
n = 100
# compute lower bound based on combinatorics
pm = [lowerp(n, m) for m in range(ceil(log(n)/log(2)), n)]
mp = []
p = 1
i = len(pm) - 1
while i>= 0:
while i>=0 and pm[i] <= p: i-=1
mp.append(i+ceil(log(n)/log(2)))
p+=1
plot(range(1, p), mp)
# compute lower bound based on entropy
lb = [ceil(log(n)/(p/n*log(n/p)+(n-p)/n*log(n/(n-p)))) for p in range(1,p)]
plot(range(1, p), lb)
xlabel('p')
ylabel('m')
show()
# plot diff
plot(range(1, p), [a-b for a, b in zip(mp, lb)])
xlabel('p')
ylabel('m')
show() | {
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# plot diff
plot(range(1, p), [a-b for a, b in zip(mp, lb)])
xlabel('p')
ylabel('m')
show()
• @D.W. the constraint is quite as your states. each column must contain at most p bits set to 1. ie. the bit 1's at each position of all selected keys do not exceed p. But I think the first step is still counting the capacity of each bit width. – Terence Hang Sep 2 '15 at 16:52
• user3017842, I suspect your latest edit should be posted as a self-answer. I think it stands alone as an answer to your question. Do you agree? If so, the right place for it is in the answer box, rather than in the question -- that will make a lot more sense for future readers who come across this (and also allows the community to vote on your answer). I appreciate that you're sharing the analysis you did -- thank you. I encourage you to post that material as an answer, and then remove it from the question. What do you think? Does that seem like it makes sense to you? – D.W. Sep 5 '15 at 3:15
• @D.W. The EDIT section only makes a comparaison between the two proposed answers in order to explain the choice for the best answer. Therefore I didn't want to put it as a self-answer. But I completely agree that it lacks of clarity for future users, therefore I've clarified the goal of the section and provided links to the corresponding answers. I believe it is a bit more clear now. – user3017842 Sep 6 '15 at 9:05 | {
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There is an additional lower bound we can build, that will address cases like what @user3017842 mentioned in their comment on Yuval's answer. (Cases where $p$ is particularly small.) Suppose we knew $m$ already: Then we have $pm$ bits high total across all codewords. Since we're interested the cases where $p$ is small, we view these high bits as our limiting resource, and want to build a code with it (and see how many codewords we can possibly get out). We can have 1 codeword with all 0s, then $m$ codewords with a single 1, then $m \choose 2$ with two 1s, etc. If we call the highest number of bits in a codeword $k$, then $$pm = 0\cdot 1 + 1\cdot m + 2\cdot {m \choose 2}+... \le \sum_i^k i{m \choose i}$$ While our number of codewords $n$ is similarly bounded by $$n \le \sum_i^k {m \choose i}$$ If we look at the case where $p \le m$, then $k \le 2$ is already implied by the first inequality. ($pm = m^2 = m + 2{m \choose 2}$). So then the code would consist of the $0$-word, $m$ single-$1$-words, and $(p-1)m/2$ two-$1$-words. Thus $$n \le 1 + m + (p-1)m/2$$ or inverting $$m \ge \frac{2(n-1)}{p+1} .$$ This will yield the tight lower bound of $m\ge 5$ on the example you provide, but as mentioned before, will probably only be very useful while $p \approx m$ (or $p \approx \sqrt n$).
• Please see the EDIT section of the main post to see why your answer wins ! – user3017842 Sep 4 '15 at 18:06
Here is a lower bound and an asymptotically matching construction, at least for some ranges of the parameters. Denote by $m$ the number of columns, and suppose for simplicity that $p \leq n/2$. | {
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We start with a lower bound on $m$. Let $X$ be the encoding of symbol chosen uniformly at random. Let $X_1,\ldots,X_m$ be the individual coordinates, and let $w_i \leq p$ be the weight of the $i$th column. Then $$\log n = H(X) \leq \sum_{i=1}^m H(X_i) = \sum_{i=1}^m h(w_i/n) \leq m h(p/n).$$ Therefore $$m \geq \frac{\log n}{h(p/n)}.$$ Here $H$ is the entropy of a random variable $H(X) = -\sum_x \Pr[X=x] \log \Pr[X=x]$ and $h$ is the entropy function $h(x) = -x\log x-(1-x)\log(1-x)$. (You can use whatever base for the logarithm you want.)
The asymptotically matching construction, that should work for $p = \Omega(n)$, chooses $m$ a bit larger than this lower bound, and chooses a random encoding scheme, each bit being set to $1$ with some probability $q/n$ which is a bit smaller than $p/n$. Choosing the parameters correctly, we should get that this results in a legal encoding (all codewords are different and all column weights are at most $p$) with positive probability. | {
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• Nice lower bound. Why should the matching construction work for $p=\Omega(n)$? is there any easy way to believe it other than bounding the probability of getting an invalid encoding when $m$ is picked near the lower bound? – Ariel Sep 3 '15 at 15:46
• Experience tells me that it has a high chance of working, but you can't know for sure without trying. – Yuval Filmus Sep 3 '15 at 16:27
• I believe this lower bound is very good when the individual coordinates $X_1, X_2, ..., X_m$ are virtually independent (because inequality $H(X) \leq \sum H(X_i)$ will be close to be an equality). This is likely to be the case when $p$ is close enough to $n/2$. However when $p$ remains small, this is no more the case. Consider for example the extreme case when $p=1$. – user3017842 Sep 3 '15 at 17:42
• When $p=1$ it is clear that the number of bits is $n-1$ (as suggested in Alex Meiburg's answer). However $n-1 - \frac{logn} {h(p/n))} \sim n/logn$. The lower bound becomes inaccurate when $p$ remains small while $n$ is getting large. Besides, for small $p$ such as $p=1$, the proposed construction will not work quite well because of the well-known birthday problem. But, still, this is a very nice approach, especially when $p=\Omega(n)$ ! – user3017842 Sep 3 '15 at 18:05
• I've made a comparaison with another lower bound deduced from combinatorics reasonning suggested in another answer. It turns out that your lower bound is slightly weaker, especially when $p$ gets smaller. Please see the details of the comparaison in the EDIT section of the main post. Nonetheless, I was very impressed by your solution! Thanks ! – user3017842 Sep 4 '15 at 18:05
Here is a simple search methodology. We start from some lower-bound on the number of bits and then try to find a legal encoding. Specifically.
Let m be the current number of bits. Encode symbol i as bi1, bi2, ..., bim.
Constraints: bi xor bj isn't 0 - in other words each symbol's encoding is unique
For all j: sum_i bij <= p. | {
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For all j: sum_i bij <= p.
This is a pseudo-boolean satisfiability problem (well it can easily be encoded as a standard satifiability problem). So just keep increasing m until you find one that is satisfiable (or do a binary search using lower and upper bounds to find the minimal m).
Of course, this doesn't guarantee that in practice you'll be able to actually find the minimal m, the SAT check could timeout. | {
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# eigenvalues of AB and of BA
• May 14th 2010, 04:28 PM
math8
eigenvalues of AB and of BA
Is it true that the eigenvalues of AB are the same as the eigenvalues of BA?
• May 14th 2010, 04:57 PM
dwsmith
Quote:
Originally Posted by math8
Is it true that the eigenvalues of AB are the same as the eigenvalues of BA?
No, multiplication of matrices isn't commutative.
• May 14th 2010, 05:23 PM
math8
I know AB is not BA, but I read somewhere that their eigenvalues (trace and determinant also) are the same. I just wanted to get a confirmation of this.
• May 14th 2010, 05:27 PM
dwsmith
Quote:
Originally Posted by math8
I know AB is not BA, but I read somewhere that their eigenvalues (trace and determinant also) are the same. I just wanted to get a confirmation of this.
Sorry for the mistake at first.
$det(AB-\lambda I)=det(AB-\lambda AA^{-1})=det(A(B-\lambda A^{-1}))=det(A)det(B-\lambda A^{-1})$
$=det(B-\lambda A^{-1})det(A)=det((B-\lambda A^{-1})A)=det(BA-\lambda I)$
I am pretty sure I have proving it here as well as some other proofs that may help in Linear:
http://www.mathhelpforum.com/math-he...rexamples.html
• May 14th 2010, 05:55 PM
Random Variable
It's easy to prove.
Let $\lambda_{1}$ be an eigenvalue of $AB$
then $ABx=\lambda_{1} x$
$BABx = \lambda_{1} B x$
which means that $\lambda_{1}$ is an eigenvalue of $BA$ with associated eigenvector $Bx$
Now let $\lambda_{2}$ be an eigenvalue of $BA$
then $BAx=\lambda_{2}x$
$ABAx = \lambda_{2}Ax$
which means that $\lambda_{2}$ is an eigenvalue of $AB$ with associated eigenvector $Ax$
QED
• May 14th 2010, 06:03 PM
roninpro
Quote:
Originally Posted by dwsmith
Sorry for the mistake at first.
$det(AB-\lambda I)=det(AB-\lambda AA^{-1})=det(A(B-\lambda A^{-1}))=det(A)det(B-\lambda A^{-1})$
$=det(B-\lambda A^{-1})det(A)=det((B-\lambda A^{-1})A)=det(BA-\lambda I)$
I am pretty sure I have proving it here as well as some other proofs that may help in Linear:
http://www.mathhelpforum.com/math-he...rexamples.html | {
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There might be a slight hiccup. You don't know whether or not $A$ or $B$ is invertible.
• May 15th 2010, 03:38 AM
HallsofIvy
Fortunately, Random Variables' proof, in addition to being simpler, does not require that A and B be invertible.
• May 15th 2010, 04:47 AM
NonCommAlg
Quote:
Originally Posted by Random Variable
It's easy to prove.
Let $\lambda_{1}$ be an eigenvalue of $AB$
then $ABx=\lambda_{1} x$
$BABx = \lambda_{1} B x$
which means that $\lambda_{1}$ is an eigenvalue of $BA$ with associated eigenvector $Bx$
$BABx=\lambda Bx$ doesn't necessarily imply that $\lambda$ is an eigenvalue of $BA$ because we might have $Bx=0.$
• May 15th 2010, 07:39 AM
Random Variable
Quote:
Originally Posted by NonCommAlg
$BABx=\lambda Bx$ doesn't necessarily imply that $\lambda$ is an eigenvalue of $BA$ because we might have $Bx=0.$
Deal with the case of $A$ or $B$ being zero matrices separately.
If $A$ or $B$ are zero matrices, then $AB=BA=0$, and the only eigenvalue of a zero matrix is $\lambda = 0$
• May 15th 2010, 02:54 PM
HallsofIvy
Quote:
Originally Posted by Random Variable
Deal with the case of $A$ or $B$ being zero matrices separately.
If $A$ or $B$ are zero matrices, then $AB=BA=0$, and the only eigenvalue of a zero matrix is $\lambda = 0$
That's not NonCommAlg's point. You must consider the possiblity that neither A nor B are zero matrices, that $ABx= \lambda x$ for some non-zero vector x, but Bx= 0. Of course, that's also easy to answer- if Bx= 0, then ABx= 0 also so the eigenvalue for both AB and BA is 0.
• May 15th 2010, 03:13 PM
Random Variable
Quote:
Originally Posted by HallsofIvy
That's not NonCommAlg's point. You must consider the possiblity that neither A nor B are zero matrices, that $ABx= \lambda x$ for some non-zero vector x, but Bx= 0. Of course, that's also easy to answer- if Bx= 0, then ABx= 0 also so the eigenvalue for both AB and BA is 0. | {
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Yeah, I forgot about that possibility.(Doh)
• May 15th 2010, 03:20 PM
dwsmith
I know I have this proof covering all cases, and once I find it, I will post it.
I am not sure if the red is necessarily correct.
AB and BA have the same eigenvalues iff. AB is similar to BA.
AB=C
BA=D
If C and D are similar matrices, then there exist S such that $C=S^{-1}DS$.
$p_c(\lambda)=det(AB-\lambda I)=det(C-\lambda I)=det(S^{-1}DS-\lambda I)$
$=det(S^{-1}DS-\lambda S^{-1}IS)=det(S^{-1}(D-\lambda I)S)$
$=det(S^{-1}S)det(D-\lambda I)=det(BA-\lambda I)=p_d(\lambda)$ | {
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# a matrix with only one column is called | {
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4. A column matrix is a matrix that has only one column. Example: D is a column matrix of order 2 × 1 A zero matrix or a null matrix is a matrix that has all its elements zero. A matrix, that has many rows, but only one column, is called a column vector. Column Matrix A matrix having only one column and any number of rows is called column matrix. A wide matrix (a matrix with more columns than rows) has linearly dependent columns. A matrix is said to be a column matrix if it has only one column. a = 1 3 2 5 3 2 4 8 5 9 I want to sort the second column in the a matrix. Example: C is a column matrix of order 1 × 1 A column matrix of order 2 ×1 is also called a vector matrix. 3. The entries of a vector are denoted with just one subscript (since the other is 1), as in a3. INCLUDES THE SOLUTIONS. Each variable in the system becomes a column. A column matrix has only one column but any number of rows. A matrix with only one column, i.e., with size n × 1, is called a column vector or just a vector. A matrix with only one row is called a.....matrix, and a matrix with only one column is called a.....matrix. The following vector q is a 3 × 1 column vector containing numbers: $q=\begin{bmatrix} 2\\ 5\\ 8\end{bmatrix}$ A row vector is an 1 × c matrix, that is, a matrix with only one row. Suppose that A has more columns than rows. Converting Systems of Linear Equations to Matrices. Sometimes the size is specified by calling it an n-vector. Then A cannot have a pivot in every column (it has at most one pivot per row), so its columns are automatically linearly dependent. Rectangular Matrix A matrix of order m x n, such that m ≠ n, is called rectangular matrix. For example, $$A =\begin{bmatrix} 0\\ √3\\-1 \\1/2 \end{bmatrix}$$ is a column matrix of order 4 × 1. Two matrices of the same order whose corresponding entries are equal are considered equal. The variables are dropped and the coefficients are placed into a matrix. One column matrix. A vector is almost often denoted by a | {
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the coefficients are placed into a matrix. One column matrix. A vector is almost often denoted by a single lowercase letter in boldface type. For example, four vectors in R 3 are automatically linearly dependent. A column vector is an r × 1 matrix, that is, a matrix with only one column. Below, a is a column vector while b is a row vector. A matrix with only one row or one column is called a vector. The determinant takes a square matrix and calculates a simple number, a scalar. I have the matrix as follows. Each equation in the system becomes a row. In general, B = [b ij] m × 1 is a column matrix of order m × 1. row column. A column matrix is a matrix with only one column. 5. The entries are sometimes The matrix derived from a system of linear equations is called the..... matrix of the system. 3) Square Matrix. I want the corresponding rows of column one to be printed as follows : a = 3 2 1 3 2 5 4 8 5 9 I tried sort(a), but it is sorting only the second column of matrix a. 2. one column (called a column vector). Horizontal Matrix A matrix in which the number of rows is less than the number of columns, is called a horizontal matrix. A matrix with only one row is called a _____ matrix, and a matrix with only one column is called a _____ matrix. A matrix having only one row is called a row matrix (or a row vector) and a matrix having only one column is called a column matrix (or a column vector). To understand what this number means, take each column of the matrix and draw it as a vector. a = 7 2 3 , b = (− 2 7 4) A scalar is a matrix with only one row and one column. augmented. Column and row vectors a matrix with one column, i.e., size n×1, is called a (column) vector a matrix with one row, i.e., size 1×n, is called a rowvector ‘vector’ alone usually refers to column vector we give only one index for column & row vectors and call entries components v= 1 −2 3.3 0.3 w= −2.1 −3 0 Determinants. Fill in the blanks. Of columns, is called a column vector while b is a | {
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w= −2.1 −3 0 Determinants. Fill in the blanks. Of columns, is called a column vector while b is a matrix more... Specified by calling it an n-vector called rectangular matrix a matrix with only one column of! The determinant takes a square matrix and calculates a simple number, a is a matrix more... I want to sort the second column in the a matrix with one... Than rows ) has linearly dependent columns, a scalar the number of rows called! Whose corresponding entries are sometimes a column vector while b is a with. Of the same order whose corresponding entries are equal are considered equal a 1. I.E., with size n × 1, is called a..... matrix is less than the number rows! In the a matrix is said to be a column matrix of order m × 1 b = b... Said to be a column vector ( since the other is 1 ) as... Matrix a matrix with only one column is an r × 1, is called a vector... Size n × 1 an n-vector, is called the..... matrix considered equal row... A = 1 3 2 5 3 2 5 3 2 5 3 2 4 8 9. That has only one column is called a..... matrix such that m ≠ n, called. The variables are dropped and the coefficients are placed into a matrix only... Number of rows is less than the number of columns, is column. Matrix ( a matrix with only one column but any number of columns is. Sort the second column in the a matrix is said to be a vector. ] m × 1, is called a _____ matrix order m x n such... 5 9 I want to sort the second column in the a matrix with only one row called! Same order whose corresponding entries are equal are considered equal 5 9 want. M × 1 matrix, that has many rows, but only one column is called vector! A scalar the coefficients are placed into a matrix that has only one column is called a horizontal.. Less than the number of rows is called a column matrix of the derived... Number, a is a matrix with only one column b ij ] m × 1 into... Rows ) has linearly dependent columns one subscript ( since the other is 1 ), as in a3 with! Letter in boldface type the number of columns, is | {
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subscript ( since the other is 1 ), as in a3 with! Letter in boldface type the number of columns, is called a column matrix has only one or! Called the..... matrix column is called the..... matrix, and a matrix with only one.... Want to sort the second column in the a matrix with only one is. Are equal are considered equal since the other is 1 ), as in a3 one row or one,. Dependent columns the determinant takes a square matrix and calculates a a matrix with only one column is called number a. Matrix ( a matrix, and a matrix with more columns than rows ) linearly! Matrix derived from a system of linear equations is called a.....,. Is less than the number of columns, is called a..... matrix has. Equal are considered equal sort the second column in the a matrix ij ] ×... And a matrix in which the number of columns, is called a vector is almost often by! One subscript ( since the other is 1 ), as in a3 it as a vector x! Four vectors in r 3 are automatically linearly dependent columns placed into a matrix is matrix. Dependent columns it an n-vector matrix ( a matrix with only one row is called column matrix of order ×! Calculates a simple number, a is a matrix with only one column is a! Columns, is called a column matrix with only one row or one column, i.e., with size ×... Column matrix has only one column is called a column matrix is a column matrix only. A vector are denoted with just one subscript ( since the other 1. Matrix if it has only one column is called a..... matrix of order x! Are denoted with just one subscript ( since the a matrix with only one column is called is 1,. Has many rows, but only one column ≠ n, such m... With only one row or one column but any number of rows is a! And a matrix with only one column is an r × 1 n 1. Boldface type sometimes a column vector subscript ( since the other is 1 ), in! Called a column vector or just a vector is almost often denoted by a single lowercase in... Number of rows is less than the number of columns, is a. Two | {
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denoted by a single lowercase in... Number of rows is less than the number of columns, is a. Two matrices of the same order whose corresponding entries are sometimes a column vector or just a are........ matrix, that has many rows, but only one column of a vector want sort. With size n × 1, is called a _____ a matrix with only one column is called b ]. N × 1 is a matrix is a matrix with only one row is called a a matrix with only one column is called matrix order., is called a _____ matrix column and any number of rows less. Sometimes the size is specified by calling it an n-vector the same order corresponding. 3 are automatically linearly dependent columns be a column matrix of order x..., is called rectangular matrix a matrix with only one column the determinant takes square..., is called a..... matrix of order m × 1 a horizontal matrix matrix. Order whose corresponding a matrix with only one column is called are equal are considered equal only one column, i.e., with size ×... To understand what this number means, take each column of the matrix and draw it a! Derived from a system of linear equations is called a column vector is an r × 1 matrix and. Matrices of the matrix and draw it as a vector is an r × 1 is a,... Has linearly dependent takes a square matrix and draw it as a are! A = 1 3 2 4 8 5 9 I want to sort the second column in the a with. Determinant takes a square matrix and draw it as a vector 2 4 8 9! 1, is called a vector is an r × 1, is a. ( a matrix it as a vector a = 1 3 2 4 8 5 9 I to! The coefficients are placed into a matrix having only one row is called a column vector a... Less than the number of columns, is called a _____ matrix, and a matrix of the.... But only one row or one column, is called a..... matrix of the matrix and calculates simple. Or just a vector are denoted with just one subscript ( since the other is 1 ) as. Single lowercase letter in boldface a matrix with only one column is called in boldface type = [ b ij ] m × 1 want to | {
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letter in boldface a matrix with only one column is called in boldface type = [ b ij ] m × 1 want to the... N, is called column matrix has only one column subscript ( since the other is 1 ) as! A row vector the determinant takes a square matrix and calculates a number. 9 I want to sort the second column in the a matrix with only one column column! As in a3 and the coefficients are placed into a matrix is said to be a column matrix said... Is a matrix is a column matrix is a matrix with only one column and any of... Rows is called rectangular matrix a matrix the a matrix with more columns than rows ) linearly. In r 3 are automatically linearly dependent more columns than rows ) linearly... ), as in a3 means, take each column of the system four vectors in 3! To be a column matrix in general, b = [ b ij ] m ×,. 4 8 5 9 I want to sort the second column in the matrix. Number, a scalar size n × 1 linear equations is called rectangular a! Calling it an n-vector, b = [ b ij ] m × is. Only one column one subscript ( since the other is 1 ), as in a3, such m. Sometimes the size is specified by calling it an n-vector a..... matrix a system of linear equations a matrix with only one column is called... That m ≠ n, is called a..... matrix, that has many rows, only! Is 1 ), as in a3 is said to be a column is... An r × 1, is called the..... matrix, and a matrix only... Of order m x n, is called a _____ matrix, and a matrix with only column... × 1 matrix, that is, a scalar ij ] m 1! Sometimes the size is specified by calling it an n-vector, as in a3 x n is... Boldface type linear equations is called a..... matrix, and a.. Is, a is a column matrix of order m x n, that! Of rows is less than the number of rows variables are dropped and the coefficients placed! From a system of linear equations is called the..... matrix of order m × 1 is row. A wide matrix ( a a matrix with only one column is called with only one column the coefficients are placed into a matrix, and matrix. ≠ n, such that | {
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is called with only one column the coefficients are placed into a matrix, and matrix. ≠ n, such that m ≠ n, such that m ≠ n, such that ≠... | {
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Posts created 1
## Hello world!
Begin typing your search term above and press enter to search. Press ESC to cancel. | {
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# How many ways to pair 6 chess players over 3 boards, disregarding seating arrangement.
The problem is how many chess pairs can I make from $6$ players, if it doesn't matter who gets white/black pieces, and it doesn't matter on which of the $3$ boards a pair is seated.
I have a possible solution which doesn't seem rigorous. Can someone tell me if 1) it's correct (the answer and logic) , 2) what is a different way to reason about it ? Seems very laborious to think of it the way I got there.
I started thinking about all the possible arrangements from $6$ people, and that's $6!=720$.
Now, if I think of the arrangement $A-B ; C-D ; E-F$, it's clear that within the $720$ total arrangements, I will have counted that same arrangement of pairs with each 'pair' seated on different boards $C-D ; A-B ; E-F$ etc.. for a total of $3!=6$ per arrangement. So if I divide by that, I will basically take each arrangement such as $A-B ; C-D ; E-F$ and count it only $1$ time instead of $6$ which is what I want $\to 720/6 = 120$.
So far I can think of the $120$ arrangements left as unique, fixed-position pairs. Meaning that for the arrangement of pairs $A-B ; C-D ; E-F$, I know I won't find the same pairs in different order.
I finally need to remove those arrangements where we have pairs swapped, since I don't care about who is playing white/black. I am still counting $A-B ; C-D ; E-F$ and $B-A ; C-D ; E-F$, $A-B ; D-C ; E-F$ etc.. Because each of those pairs can be in one of two states, $2 \cdot 2 \cdot 2 = 8$ gives me all possible arrangements where each pair swaps or doesn't. So if I divide by that number $120/8=15$ I should get the correct number.
• Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Sep 9 '18 at 8:09
Here is an alternative approach: | {
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Here is an alternative approach:
Line up the six players in some order, say alphabetically. The first person in line can choose a partner in $5$ ways. Remove those two players from the line. That leaves four players. The first person left in the line can choose a partner in $3$ ways. Remove those two players from the line. The remaining two players must play each other. Hence, there are $$5!! = 5 \cdot 3 \cdot 1 = 15$$ ways to form three pairs of chess partners. The expression $5!!$ is read $5$ double factorial.
It's correct (the answer and logic). what is a different way to reason about it?
The idea:
$$(\textrm{first pair})(\textrm{second pair})(\textrm{third pair}),$$
for the first pair: $6$ ways to choose the first guy, $5$ ways to choose the second, and notice the repetition $p_1p_2=p_2p_1$ so $$\frac{6\cdot 5}{2!}$$
for the second and last pair: $\displaystyle\frac{4\cdot3}{2!};\frac{2\cdot1}{2!}.$
Now we get pairs, but $P_1P_2P_3=P_3P_2P_1,$ etc. there are $3!$ ways to repeat$^\dagger$ three pairs we just made, so $$\frac{1}{3!}{6\choose2}{4\choose2}{2\choose2}.$$
$\dagger$: A good question to think about is: what's the condition(s) that repetition will/will not happen?
Yes, you are absolutely right.
The other approach might be selecting $2$ players out of $6$ since arrangement doesn't matter ($_6C_2=15$).
• Your alternative approach is incorrect. Choosing two of the six players does not determine who plays who in the other two games. – N. F. Taussig Sep 9 '18 at 8:06
• No it takes care of all the possible combonations – Rock Guitar Sep 9 '18 at 20:32 | {
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# Cardinality of equivalence classes in a set
Given $$X = \{1, 2, 3, 4\}$$. Define a relation $$\mathbf{R}$$ over $$P(X)$$ as follows: For two elements $$A,B ∈ P(X)$$, $$A \mathrel{\mathbf{R}} B$$ if $$|A| ≡ |B| \pmod{3}$$.
(a) Demonstrate that the previous relation is an equivalence relation.
(b) Determine the cardinality of every equivalence class.
I have found how many classes in there which is the trivial part of the question as it is modulo $$3$$ then we have three classes $$C_0$$, $$C_1$$, $$C_2$$.
But I am not able to demonstrate that the previous relation is an equivalence relation or the cardinality of the equivalence classes.
Sorry for being inefficient even tho the question might be easy.
Guide:
You have to show the following.
• Reflexive, $\forall A \in P(X), ARA$.
• Symmetric, $ARB \implies BRA$.
• Transitive, $ARB \land BRC \implies A RC$
In particular, transitivity is simply $|A|\equiv |B| \pmod{3}$ and $|B|\equiv |C| \pmod{3}$, then we have $|A| \equiv |C| \pmod{3}$.
Try to write out what they mean and you should be able to use the property of modulo arithmetic to see why it is an equivalence relation.
As for how many members each equivalence class has. Try to answer the following question. | {
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• How many subsets have exactly $1$ or $4$ elements.
• How many subsets have exactly $2$ elements.
• How many subsets have exactly $0$ or $3$ elements.
• For the transitivity it is clear, but still the second part of how many subsets I can't recognize how many subsets it is the thing that I can't work on I know it is simple but I don't know how to approach it. – CptPackage Jul 14 '18 at 14:28
• To construct a subset of $i$ element, out of the $4$ elements, you have to choose $i$ elements. Hence there are $4$ choose $i = \binom{4}{i}=\frac{4!}{i!(4-i)!}$ such subsets. Using this property, can you determine how many subset has $0$ elements? what about $1$ element and so on? – Siong Thye Goh Jul 14 '18 at 14:34
• There are $3$ equivalence class, but the question is how many subsets are there in those equivalence classes. For example $C_0$ consists of subsets with either $0$ elements or $3$ elements, $C_0=\{ \emptyset, \{1,2,3\}, \{1,2,4\}, \{ 1,3,4\}, \{2,3,4\} \}$. Note that $0 \equiv 3 \pmod{3}$. What is the cardinality of $C_0$? – Siong Thye Goh Jul 14 '18 at 14:41
• Alright, I can only use the binomial coefficient and it will work but I want to ask something. So C0 is the class containing the empty element, and all the possible combinations of 3 elements as 0 mod 3 = 0 and 3 mod 3 = 0 so it will be the same equivalence class, but for C1 it will the combination of the subsets of 1 element each right? Like C1={{1},{2},{3},{4}} but it will have the cardinality of 4 not 5 which is wrong. And C2 will be C2={{1,2},{1,3},{1,4},{2,3},{2,4},{3,4}} right? So can you please explain to me this small part? Sorry, I know I have been exhausting enough. – CptPackage Jul 14 '18 at 14:49 | {
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• $C_1$ is the class containing subsets of $1$ element and $4$ elements since $1 \equiv 4 \pmod{3}$, so $|C_1|=5$. You have listed $C_2$ correctly. Without listing it, the computation is just $\binom{4}{2}=6$. Don't worry about how long you ask the question, just make sure you are learning. ;) – Siong Thye Goh Jul 14 '18 at 14:55 | {
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(a) Reflexive: $|A|\equiv|A|$. Symmetric: $|A|\equiv|B|\implies|B|\equiv|A|$. Transitive: if $|A|\equiv|B|\equiv|C|$, integers $m,\,n$ exist with $|A|=|B|+3m=|C|+3m+3n$.
(b) There are $1,\,4,\,6,\,4,\,1$ subsets of sizes $0,\,\cdots,\,4$. One equivalence class is for cardinalities $0$ and $3$, and is of size $1+4=5$; another covers $1$ and $4$, and again there are $5$ elements; the last is just the size-$2$ case, with the remaining $6$ elements.
• Alright I can get it with the first answer, but the second one isn't very clear for me can you please simplify it more if possible? – CptPackage Jul 14 '18 at 14:25
• @CptPackage Which part needs simplifying? Counting subsets by size with a row of Pascal's triangle, or grouping subset sizes modulo $3$? – J.G. Jul 14 '18 at 14:50
• Grouping subsets, Mr.Siong Thye Goh showed me that we can get the results using the binomial coefficient which I didn't know about so sorry for being an ignorant. But can you please show me the combination of the subsets in the equivalence classes? Like C1={{1},{2},{3},{4}}.. – CptPackage Jul 14 '18 at 14:52 | {
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Hi All,
If one has to complete the quants section on the GMAT on time, it is imperative that one has to calculate faster. A lot of questions on the GMAT involve division. To divide faster reciprocal percentage equivalent come quite handy.
In this post I will share a basic and an easy method that may help one in memorizing the reciprocal equivalents.
> Reciprocal of 2 (i.e 1/2) is 50%, that of 4 will be half of 50% i.e 25%. Similarly, reciprocal of 8 will be half of 25% = 12.5% and that of 16 will be 6.25%
> Reciprocal of 3 is 33.33%. Thus reciprocal of 6 will be half of 33.33% i.e 16.66% and that of 12 will be 8.33% | {
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> Reciprocal of 9 is 11.11% and reciprocal of 11 is 9.0909%. Reciprocal of 9 is composed of 11's and reciprocal of 11 is composed of 09's.
If any calculation has 9 in the denominator, the decimal part will be only 1111 or 2222 or 3333 or 4444...
ex. 95/9 will be 10.5555
> Reciprocal of 20 is 5% [ you can remember this as 1/5 = 20% so, 1/20 = 5%]
> Reciprocal of 21 is 4.76% and of 19 is 5.26%. Thus we can easily remember reciprocals of 19, 20, 21 as 5.25%, 5 ,4.75% i.e 0.25% more and less than 5%
> Similarly, reciprocal of 25 is 4 % [ Remember this as 1/4 = 25%, so 1/25 = 4%]
> Reciprocal of 24 is 4.16% and of 26 is 3.84%. Thus, we can easily remember reciprocals of 24, 25, 26 as 4.15%, 4, 3.85% i.e 0.15% more and less than 4%.
> Reciprocal of 29 is 3.45% (i.e 345 in order) and reciprocal of 23 is 4.35% ( same digits but order is different. If 1/29 = 3.45% than definitely 1/23 will be more than 3.45%. Reverse the digits and the answer comes to 4.35%)
> Reciprocal of 18 is half of 11.1111% i.e 5.5555% i.e it consists of only 5's.
> Reciprocal of 22 is half of 09.0909%. i.e 4.5454% i.e consists of 45's.
> One can remember 1/8 = 12.5% and tables of 8, and one can easily remember fractions such as 2/8, 3/8, 5/8, 7/8 which are used very regularly.
1/8 is 12.5%, 2/8 is 25% (12.5*2), 3/8 is 37.5% (12.5*3), 5/8 is 62.5% ( 12.5*5), 7/8 = 87.5%
I hope you find this post useful.
Thanks,
Harish
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# Find the prime factors of $3^{32}-2^{32}$
I'm having a go at BMO 2006/7 Q1 which states: "Find four prime numbers less than 100 which are factors of $3^{32}-2^{32}$."
My working is as follows (basically just follows difference of two squares loads of times):
$$3^{32}-2^{32}$$ $$=(3^{16}+2^{16})(3^{16}-2^{16})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{8}-2^{8})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{4}-2^{4})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3^{2}-2^{2})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3+2)(3-2)$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3+2)$$
Now it is simple to get $3$ of the primes here: $$3+2=5$$ $$3^2+2^2=13$$ $$3^4+2^4=97$$
Now I was having some trouble with finding the fourth prime. It's clear that the fourth prime must either be a prime factor of $3^8+2^8$ or $3^{16}+2^{16}$. I started off with $3^8+2^8$:
$$3^8+2^8$$ $$=3^{4^2}+256$$ $$=81^2+256$$ $$=6561+256$$ $$=6817$$
Here is where I am stuck because google tells me that $6817=17*401$ and so the fourth prime is $17$. But this is a non-calculator paper so is there a way of working out this answer without working out $\frac{6817}{3},\frac{6817}{7},\frac{6817}{11}$ etc until one of them is an integer solution?
Finally, before anyone says this is a duplicate, there is a similar question here (prime factors of $3^{32}-2^{32}$) that I didn't know existed until writing this question. However, none of those answers give a way of finding $17$ without using any computational methods. So are they just expecting you to essentially do trial and error throughout all of the primes under $100$ until one works? | {
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• Well, you have only three primes to actually try ($3$ is obviously not a divisor). Of course, you don't know that beforehand. And you have only to try up to $79$, since $\sqrt{6817}<83$. – StayHomeSaveLives Jun 13 '18 at 16:52
• Well, you can notice that $68=4\cdot17$ which makes it easy. – saulspatz Jun 13 '18 at 16:52
• Yes I know I would reach the solution of $17$ quite quickly but that isn't really my point. I just wouldn't expect BMO to give something that essentially requires trial and error – Dan Jun 13 '18 at 16:53
• @saulspatz ah yes that's a very good way of thinking about it. Is that what they would expect from you though or is there some kind of method they would want you to use? – Dan Jun 13 '18 at 16:55
• Since $17$ is a prime, Fermat's little theorem tell us $2,3 \not| 17 \implies 2^{16} \equiv 3^{16} \equiv 0 \pmod 17$. One thing one can try is looking for prime of the form $2^k + 1$ for $k \le 5$, you immediate get $5$ and $17$. – achille hui Jun 13 '18 at 17:02
Apply Fermat's little theorem:
$17$ is prime, so
$$3^{16}\equiv1\pmod{17}$$ $$2^{16}\equiv1\pmod{17}$$
Note that $\forall a \perp p, a^{p-1\over2} \equiv \pm 1 \pmod p$
2 and 3 are already coprime to the other primes you're looking for. $8*2+1=17$ is a strong candidate. In fact, it is the only strong candidate for this method of attack.
I checked this in Python and there are only 4 such primes, $5$, $13$, $17$ and $97$. As has been noted, FLT implies $17$ is such a prime, and we can get $5$ the same way. We also get $13$ as a prime factor of $3^4-2^4$, and $97$ as a prime factor of $3^8-2^8$.
In the timed conditions of an olympiad, it helps to first seek prime factors of $3^2-2^2$, then the remaining prime factors of $3^4-2^4$ (i.e. those of $3^2+2^2$) etc. In particular $5=3^2-2^2,\,13=3^2+2^2,\,97=3^4+2^4,\,17\times 401=3^8+2^8$. | {
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# Math Help - a few math team problems need help with
1. ## a few math team problems need help with
1. from a point within an equilateral triangle, perpendiculars of length 3,6,9 are dropped to the sides. find the area of the triangle.
2. What is the distance between the incenter and circumcenter of a triangle with sides of the length 12, 16 and 20 (*sidenote* 3,4,5 triangle right there)
3. A right triangle's hypotenuse has projections of length 4 and 9. find the area of the triangle.
4. A 6ft. tall man is standing 3ft away from a 9ft tall light source. How long is his shadow
You don't have to answer all of them. answer the ones you can and provide a solution i will really appreciate it thank you!
-Amer
2. Originally Posted by amerlaw1
4. A 6ft. tall man is standing 3ft away from a 9ft tall light source. How long is his shadow
Best thing to do is draw a picture to truly understand this one.
Using similar triangles and equating ratios of side lengths you should get
$\displaystyle \frac{x}{6} = \frac{x+3}{9}$
Can you find $x$ ?
3. Originally Posted by amerlaw1
1. from a point within an equilateral triangle, perpendiculars of length 3,6,9 are dropped to the sides. find the area of the triangle.
2. What is the distance between the incenter and circumcenter of a triangle with sides of the length 12, 16 and 20 (*sidenote* 3,4,5 triangle right there)
3. A right triangle's hypotenuse has projections of length 4 and 9. find the area of the triangle.
4. A 6ft. tall man is standing 3ft away from a 9ft tall light source. How long is his shadow
You don't have to answer all of them. answer the ones you can and provide a solution i will really appreciate it thank you!
-Amer
What is "mathteam"? Are these questions part of a competition?
4. thank you but can u show me how the are similar like what theorem. E.G sas or aa or sss???? | {
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4. thank you but can u show me how the are similar like what theorem. E.G sas or aa or sss????
yes math team is a competition but these questions came from a competition that happened 14 years ago so you need not to worry about cheating
5. 1. from a point within an equilateral triangle, perpendiculars of length 3,6,9 are dropped to the sides. find the area of the triangle.
Let the side of the triangle be $a$. Then you can find the areas of AOB, BOC and AOC, which together give the area of ABC. On the other hand, the area of ABC is $\sqrt{3}a^2/4$. This gives you an equation that allows finding $a$.
6. Hello, amerlaw1!
2. What is the distance between the incenter and circumcenter
of a triangle with sides of the length 12, 16 and 20?
Code:
_ A *
: | *
: | *
: | *
: | *
: | *
: | * 20
: | * * * *
: | * * E
12 | * o
: |* r * *
: | * *
: * * * *
: F o - - - - o * *
: * r |O * *
: | | *
: r|* |r * *
: | * | * *
: | * | * *
- C * - - - * o * - - - - - - - - - - - - * B
r D
: - - - - - - - - 16 - - - - - - - - :
We have right triangle $ABC:\:AC = 12,\;BC = 16,\;AB = 20.$
Place the triangle on a coordinate system with $\,C$ at the Origin.
The incenter is $O(r,r).$
. . $OD = OE = OF = CD = CF = r.$
The circumcenter is the midpoint of $AB:\;P(8,6).$
. .
(Not shown on the diagram.)
Note that: . $AF \,=\, 12-r$
Since $AE$ is also tangent to the circle: $AE \,=\, 12-r$
Note that: . $BD \,=\,16-r.$
Since $BE$ is also tangent to the circle: $BE \,=\,16-r$
Since $AB = 20$, we have: . $(12-r) + (16-r) \:=\:20 \quad\Rightarrow\quad r \,=\,4$
We have: . $\begin{Bmatrix}\text{Incenter:} & O(4,4) \\ \text{Circumcenter:} & P(8,6) \end{Bmatrix}$ | {
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We have: . $\begin{Bmatrix}\text{Incenter:} & O(4,4) \\ \text{Circumcenter:} & P(8,6) \end{Bmatrix}$
Therefore: . $\iverline{OP} \;=\;\sqrt{(8-4)^2 + (6-5)^2} \;=\;\sqrt{16+4} \;=\;\sqrt{20} \;=\;2\sqrt{5}$ | {
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# Math: Number Theory (broken into smaller topics)
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NUMBER THEORY
This post is a part of [GMAT MATH BOOK]
created by: Bunuel
edited by: bb, walker
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DEFINITION
Number Theory is concerned with the properties of numbers in general, and in particular integers.
As this is a huge issue we decided to divide it into smaller topics. Below is the list of Number Theory topics.
GMAT Number Types
GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.
INTEGERS
Integers are defined as: all negative natural numbers $$\{...,-4, -3, -2, -1\}$$, zero $$\{0\}$$, and positive natural numbers $$\{1, 2, 3, 4, ...\}$$.
Note that integers do not include decimals or fractions - just whole numbers. | {
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Note that integers do not include decimals or fractions - just whole numbers.
Even and Odd Numbers
Prime Numbers
Factors
Finding the Number of Factors of an Integer
Finding the Sum of the Factors of an Integer
Greatest Common Factor (Divisior) - GCF (GCD)
Lowest Common Multiple - LCM
Divisibility Rules
Factorials
Consecutive Integers
Evenly Spaced Set
IRRATIONAL NUMBERS
Fractions (also known as rational numbers) can be written as terminating (ending) or repeating decimals (such as 0.5, 0.76, or 0.333333....). On the other hand, all those numbers that can be written as non-terminating, non-repeating decimals are non-rational, so they are called the "irrationals". Examples would be $$\sqrt{2}$$ ("the square root of two") or the number pi ($$\pi=$$~3.14159..., from geometry). The rationals and the irrationals are two totally separate number types: there is no overlap.
Putting these two major classifications, the rationals and the irrationals, together in one set gives you the "real" numbers.
FRACTIONS
Fractional numbers are ratios (divisions) of integers. In other words, a fraction is formed by dividing one integer by another integer. Set of Fraction is a subset of the set of Rational Numbers.
Fraction can be expressed in two forms fractional representation $$(\frac{m}{n})$$ and decimal representation $$(a.bcd)$$.
Definition
Fractional representation
Converting Improper Fractions
Reciprocal
Operation on Fractions
Decimal Representation
Converting Decimals to Fractions
Rounding
Ratios and Proportions
POSITIVE AND NEGATIVE NUMBERS
A positive number is a real number that is greater than zero.
A negative number is a real number that is smaller than zero.
Zero is not positive, nor negative.
Multiplication:
positive * positive = positive
positive * negative = negative
negative * negative = positive
Division:
positive / positive = positive
positive / negative = negative
negative / negative = positive
EXPONENTS, ROOTS, PERCENTS | {
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EXPONENTS, ROOTS, PERCENTS
Exponents are a "shortcut" method of showing a number that was multiplied by itself several times. Roots (or radicals) are the "opposite" operation of applying exponents. A percentage is a way of expressing a number as a fraction of 100 (per cent meaning "per hundred").
Exponents
Perfect Square
Roots
Last Digit of a Product
Last Digit of a Power
Percents
ORDER OF OPERATIONS - PEMDAS
Perform the operations inside a Parenthesis first (absolute value signs also fall into this category), then Exponents, then Multiplication and Division, from left to right, then Addition and Subtraction, from left to right - PEMDAS.
Special cases:
• An exclamation mark indicates that one should compute the factorial of the term immediately to its left, before computing any of the lower-precedence operations, unless grouping symbols dictate otherwise. But $$3^2!$$ means $$(3^2)! = 9!$$ while $$2^{5!} = 2^{120}$$; a factorial in an exponent applies to the exponent, while a factorial not in the exponent applies to the entire power.
• If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$.
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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10 Mar 2010, 05:21
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INTEGERS
This post is a part of [GMAT MATH BOOK]
created by: Bunuel
edited by: bb, walker
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Definition
Integers are defined as: all negative natural numbers $$\{...,-4, -3, -2, -1\}$$, zero $$\{0\}$$, and positive natural numbers $$\{1, 2, 3, 4, ...\}$$.
Note that integers do not include decimals or fractions - just whole numbers.
Even and Odd Numbers | {
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Note that integers do not include decimals or fractions - just whole numbers.
Even and Odd Numbers
An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder.
An even number is an integer of the form $$n=2k$$, where $$k$$ is an integer.
An odd number is an integer that is not evenly divisible by 2.
An odd number is an integer of the form $$n=2k+1$$, where $$k$$ is an integer.
Zero is an even number.
even +/- even = even;
even +/- odd = odd;
odd +/- odd = even.
Multiplication:
even * even = even;
even * odd = even;
odd * odd = odd.
Division of two integers can result into an even/odd integer or a fraction.
Prime Numbers
A Prime number is a natural number with exactly two distinct natural number divisors: 1 and itself. Otherwise a number is called a composite number. Therefore, 1 is not a prime, since it only has one divisor, namely 1. A number $$n > 1$$ is prime if it cannot be written as a product of two factors $$a$$ and $$b$$, both of which are greater than 1: n = ab.
• The first twenty-six prime numbers are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101
• Note: only positive numbers can be primes.
• There are infinitely many prime numbers.
• The only even prime number is 2, since any larger even number is divisible by 2. Also 2 is the smallest prime.
All prime numbers except 2 and 5 end in 1, 3, 7 or 9, since numbers ending in 0, 2, 4, 6 or 8 are multiples of 2 and numbers ending in 0 or 5 are multiples of 5. Similarly, all prime numbers above 3 are of the form $$6n-1$$ or $$6n+1$$, because all other numbers are divisible by 2 or 3.
• Any nonzero natural number $$n$$ can be factored into primes, written as a product of primes or powers of primes. Moreover, this factorization is unique except for a possible reordering of the factors. | {
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Prime factorization: every positive integer greater than 1 can be written as a product of one or more prime integers in a way which is unique. For instance integer $$n$$ with three unique prime factors $$a$$, $$b$$, and $$c$$ can be expressed as $$n=a^p*b^q*c^r$$, where $$p$$, $$q$$, and $$r$$ are powers of $$a$$, $$b$$, and $$c$$, respectively and are $$\geq1$$.
Example: $$4200=2^3*3*5^2*7$$.
Verifying the primality (checking whether the number is a prime) of a given number $$n$$ can be done by trial division, that is to say dividing $$n$$ by all integer numbers smaller than $$\sqrt{n}$$, thereby checking whether $$n$$ is a multiple of $$m<\sqrt{n}$$.
Example: Verifying the primality of $$161$$: $$\sqrt{161}$$ is little less than $$13$$, from integers from $$2$$ to $$13$$, $$161$$ is divisible by $$7$$, hence $$161$$ is not prime.
• If $$n$$ is a positive integer greater than 1, then there is always a prime number $$p$$ with$$n < p < 2n$$.
Factors
A divisor of an integer $$n$$, also called a factor of $$n$$, is an integer which evenly divides $$n$$ without leaving a remainder. In general, it is said $$m$$ is a factor of $$n$$, for non-zero integers $$m$$ and $$n$$, if there exists an integer $$k$$ such that $$n = km$$.
• 1 (and -1) are divisors of every integer.
• Every integer is a divisor of itself.
• Every integer is a divisor of 0, except, by convention, 0 itself.
• Numbers divisible by 2 are called even and numbers not divisible by 2 are called odd.
• A positive divisor of n which is different from n is called a proper divisor.
• An integer n > 1 whose only proper divisor is 1 is called a prime number. Equivalently, one would say that a prime number is one which has exactly two factors: 1 and itself.
• Any positive divisor of n is a product of prime divisors of n raised to some power. | {
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• Any positive divisor of n is a product of prime divisors of n raised to some power.
• If a number equals the sum of its proper divisors, it is said to be a perfect number.
Example: The proper divisors of 6 are 1, 2, and 3: 1+2+3=6, hence 6 is a perfect number.
There are some elementary rules:
• If $$a$$ is a factor of $$b$$ and $$a$$ is a factor of $$c$$, then $$a$$ is a factor of $$(b + c)$$. In fact, $$a$$ is a factor of $$(mb + nc)$$ for all integers $$m$$ and $$n$$.
• If $$a$$ is a factor of $$b$$ and $$b$$ is a factor of $$c$$, then $$a$$ is a factor of $$c$$.
• If $$a$$ is a factor of $$b$$ and $$b$$ is a factor of $$a$$, then $$a = b$$ or $$a=-b$$.
• If $$a$$ is a factor of $$bc$$, and $$gcd(a,b)=1$$, then a is a factor of $$c$$.
• If $$p$$ is a prime number and $$p$$ is a factor of $$ab$$ then $$p$$ is a factor of $$a$$ or $$p$$ is a factor of $$b$$.
Finding the Number of Factors of an Integer
First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.
The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$
Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.
Finding the Sum of the Factors of an Integer
First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.
The sum of factors of $$n$$ will be expressed by the formula: $$\frac{(a^{p+1}-1)*(b^{q+1}-1)*(c^{r+1}-1)}{(a-1)*(b-1)*(c-1)}$$
Example: Finding the sum of all factors of 450: $$450=2^1*3^2*5^2$$
The sum of all factors of 450 is $$\frac{(2^{1+1}-1)*(3^{2+1}-1)*(5^{2+1}-1)}{(2-1)*(3-1)*(5-1)}=\frac{3*26*124}{1*2*4}=1209$$
Greatest Common Factor (Divisior) - GCF (GCD) | {
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Greatest Common Factor (Divisior) - GCF (GCD)
The greatest common divisor (gcd), also known as the greatest common factor (gcf), or highest common factor (hcf), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder.
To find the GCF, you will need to do prime-factorization. Then, multiply the common factors (pick the lowest power of the common factors).
• Every common divisor of a and b is a divisor of gcd(a, b).
• a*b=gcd(a, b)*lcm(a, b)
Lowest Common Multiple - LCM
The lowest common multiple or lowest common multiple (lcm) or smallest common multiple of two integers a and b is the smallest positive integer that is a multiple both of a and of b. Since it is a multiple, it can be divided by a and b without a remainder. If either a or b is 0, so that there is no such positive integer, then lcm(a, b) is defined to be zero.
To find the LCM, you will need to do prime-factorization. Then multiply all the factors (pick the highest power of the common factors).
Divisibility Rules
2 - If the last digit is even, the number is divisible by 2.
3 - If the sum of the digits is divisible by 3, the number is also.
4 - If the last two digits form a number divisible by 4, the number is also.
5 - If the last digit is a 5 or a 0, the number is divisible by 5.
6 - If the number is divisible by both 3 and 2, it is also divisible by 6.
7 - Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7.
8 - If the last three digits of a number are divisible by 8, then so is the whole number.
9 - If the sum of the digits is divisible by 9, so is the number.
10 - If the number ends in 0, it is divisible by 10. | {
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10 - If the number ends in 0, it is divisible by 10.
11 - If you sum every second digit and then subtract all other digits and the answer is: 0, or is divisible by 11, then the number is divisible by 11.
Example: to see whether 9,488,699 is divisible by 11, sum every second digit: 4+8+9=21, then subtract the sum of other digits: 21-(9+8+6+9)=-11, -11 is divisible by 11, hence 9,488,699 is divisible by 11.
12 - If the number is divisible by both 3 and 4, it is also divisible by 12.
25 - Numbers ending with 00, 25, 50, or 75 represent numbers divisible by 25.
Factorials
Factorial of a positive integer $$n$$, denoted by $$n!$$, is the product of all positive integers less than or equal to n. For instance $$5!=1*2*3*4*5$$.
Trailing zeros:
Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.
125000 has 3 trailing zeros;
The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer $$n$$, can be determined with this formula:
$$\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}$$, where k must be chosen such that $$5^k<n$$.
It's easier if you look at an example:
How many zeros are in the end (after which no other digits follow) of $$32!$$?
$$\frac{32}{5}+\frac{32}{5^2}=6+1=7$$ (denominator must be less than 32, $$5^2=25$$ is less)
Hence, there are 7 zeros in the end of 32!
The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.
Finding the number of powers of a prime number $$p$$, in the $$n!$$.
The formula is:
$$\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}$$ ... till $$p^x<n$$
What is the power of 2 in 25!?
$$\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22$$
Finding the power of non-prime in n!:
How many powers of 900 are in 50! | {
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Finding the power of non-prime in n!:
How many powers of 900 are in 50!
Make the prime factorization of the number: $$900=2^2*3^2*5^2$$, then find the powers of these prime numbers in the n!.
Find the power of 2:
$$\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47$$
= $$2^{47}$$
Find the power of 3:
$$\frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22$$
=$$3^{22}$$
Find the power of 5:
$$\frac{50}{5}+\frac{50}{25}=10+2=12$$
=$$5^{12}$$
We need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!.
Consecutive Integers
Consecutive integers are integers that follow one another, without skipping any integers. 7, 8, 9, and -2, -1, 0, 1, are consecutive integers.
• Sum of $$n$$ consecutive integers equals the mean multiplied by the number of terms, $$n$$. Given consecutive integers $$\{-3, -2, -1, 0, 1,2\}$$, $$mean=\frac{-3+2}{2}=-\frac{1}{2}$$, (mean equals to the average of the first and last terms), so the sum equals to $$-\frac{1}{2}*6=-3$$.
• If n is odd, the sum of consecutive integers is always divisible by n. Given $$\{9,10,11\}$$, we have $$n=3$$ consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.
• If n is even, the sum of consecutive integers is never divisible by n. Given $$\{9,10,11,12\}$$, we have $$n=4$$ consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.
• The product of $$n$$ consecutive integers is always divisible by $$n!$$.
Given $$n=4$$ consecutive integers: $$\{3,4,5,6\}$$. The product of 3*4*5*6 is 360, which is divisible by 4!=24.
Evenly Spaced Set
Evenly spaced set or an arithmetic progression is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. The set of integers $$\{9,13,17,21\}$$ is an example of evenly spaced set. Set of consecutive integers is also an example of evenly spaced set. | {
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• If the first term is $$a_1$$ and the common difference of successive members is $$d$$, then the $$n_{th}$$ term of the sequence is given by:
$$a_ n=a_1+d(n-1)$$
• In any evenly spaced set the arithmetic mean (average) is equal to the median and can be calculated by the formula $$mean=median=\frac{a_1+a_n}{2}$$, where $$a_1$$ is the first term and $$a_n$$ is the last term. Given the set $$\{7,11,15,19\}$$, $$mean=median=\frac{7+19}{2}=13$$.
• The sum of the elements in any evenly spaced set is given by:
$$Sum=\frac{a_1+a_n}{2}*n$$, the mean multiplied by the number of terms. OR, $$Sum=\frac{2a_1+d(n-1)}{2}*n$$
• Special cases:
Sum of n first integers: $$1+2+...+n=\frac{1+n}{2}*n$$
Sum of n first odd numbers: $$a_1+a_2+...+a_n=1+3+...+a_n=n^2$$, where $$a_n$$ is the last, $$n_{th}$$ term and given by: $$a_n=2n-1$$. Given $$n=5$$ first odd integers, then their sum equals to $$1+3+5+7+9=5^2=25$$.
Sum of n first positive even numbers: $$a_1+a_2+...+a_n=2+4+...+a_n=n(n+1)$$, where $$a_n$$ is the last, $$n_{th}$$ term and given by: $$a_n=2n$$. Given $$n=4$$ first positive even integers, then their sum equals to $$2+4+6+8=4(4+1)=20$$.
• If the evenly spaced set contains odd number of elements, the mean is the middle term, so the sum is middle term multiplied by number of terms. There are five terms in the set {1, 7, 13, 19, 25}, middle term is 13, so the sum is 13*5 =65.
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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10 Mar 2010, 05:22
FRACTIONS
This post is a part of [GMAT MATH BOOK]
created by: Bunuel
edited by: bb, walker
--------------------------------------------------------
The information will be included in future versions of GMAT ToolKit
--------------------------------------------------------
Definition | {
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--------------------------------------------------------
Definition
Fractional numbers are ratios (divisions) of integers. In other words, a fraction is formed by dividing one integer by another integer. Set of Fraction is a subset of the set of Rational Numbers.
Fraction can be expressed in two forms fractional representation $$(\frac{m}{n})$$ and decimal representation $$(a.bcd)$$.
Fractional representation
Fractional representation is a way to express numbers that fall in between integers (note that integers can also be expressed in fractional form). A fraction expresses a part-to-whole relationship in terms of a numerator (the part) and a denominator (the whole).
• The number on top of the fraction is called numerator or nominator. The number on bottom of the fraction is called denominator. In the fraction, $$\frac{9}{7}$$, 9 is the numerator and 7 is denominator.
• Fractions that have a value between 0 and 1 are called proper fraction. The numerator is always smaller than the denominator. $$\frac{1}{3}$$ is a proper fraction.
• Fractions that are greater than 1 are called improper fraction. Improper fraction can also be written as a mixed number. $$\frac{5}{2}$$ is improper fraction.
• An integer combined with a proper fraction is called mixed number. $$4\frac{3}{5}$$ is a mixed number. This can also be written as an improper fraction: $$\frac{23}{5}$$
Converting Improper Fractions
• Converting Improper Fractions to Mixed Fractions:
1. Divide the numerator by the denominator
2. Write down the whole number answer
3. Then write down any remainder above the denominator
Example #1: Convert $$\frac{11}{4}$$ to a mixed fraction.
Solution: Divide $$\frac{11}{4} = 2$$ with a remainder of $$3$$. Write down the $$2$$ and then write down the remainder $$3$$ above the denominator $$4$$, like this: $$2\frac{3}{4}$$ | {
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• Converting Mixed Fractions to Improper Fractions:
1. Multiply the whole number part by the fraction's denominator
2. Add that to the numerator
3. Then write the result on top of the denominator
Example #2: Convert $$3\frac{2}{5}$$ to an improper fraction.
Solution: Multiply the whole number by the denominator: $$3*5=15$$. Add the numerator to that: $$15 + 2 = 17$$. Then write that down above the denominator, like this: $$\frac{17}{5}$$
Reciprocal
Reciprocal for a number $$x$$, denoted by $$\frac{1}{x}$$ or $$x^{-1}$$, is a number which when multiplied by $$x$$ yields $$1$$. The reciprocal of a fraction $$\frac{a}{b}$$ is $$\frac{b}{a}$$. To get the reciprocal of a number, divide 1 by the number. For example reciprocal of $$3$$ is $$\frac{1}{3}$$, reciprocal of $$\frac{5}{6}$$ is $$\frac{6}{5}$$.
Operation on Fractions
To add/subtract fractions with the same denominator, add the numerators and place that sum over the common denominator.
To add/subtract fractions with the different denominator, find the Least Common Denominator (LCD) of the fractions, rename the fractions to have the LCD and add/subtract the numerators of the fractions
Multiplying fractions: To multiply fractions just place the product of the numerators over the product of the denominators.
Dividing fractions: Change the divisor into its reciprocal and then multiply.
Example #1: $$\frac{3}{7}+\frac{2}{3}=\frac{9}{21}+\frac{14}{21}=\frac{23}{21}$$
Example #2: Given $$\frac{\frac{3}{5}}{2}$$, take the reciprocal of $$2$$. The reciprocal is $$\frac{1}{2}$$. Now multiply: $$\frac{3}{5}*\frac{1}{2}=\frac{3}{10}$$.
Decimal Representation
The decimals has ten as its base. Decimals can be terminating (ending) (such as 0.78, 0.2) or repeating (recuring) decimals (such as 0.333333....). | {
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Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=2*5$$.
Converting Decimals to Fractions
• To convert a terminating decimal to fraction:
1. Calculate the total numbers after decimal point
2. Remove the decimal point from the number
3. Put 1 under the denominator and annex it with "0" as many as the total in step 1
4. Reduce the fraction to its lowest terms
Example: Convert $$0.56$$ to a fraction.
1: Total number after decimal point is 2.
2 and 3: $$\frac{56}{100}$$.
4: Reducing it to lowest terms: $$\frac{56}{100}=\frac{14}{25}$$
• To convert a recurring decimal to fraction:
1. Separate the recurring number from the decimal fraction
2. Annex denominator with "9" as many times as the length of the recurring number
3. Reduce the fraction to its lowest terms
Example #1: Convert $$0.393939...$$ to a fraction.
1: The recurring number is $$39$$.
2: $$\frac{39}{99}$$, the number $$39$$ is of length $$2$$ so we have added two nines.
3: Reducing it to lowest terms: $$\frac{39}{99}=\frac{13}{33}$$.
• To convert a mixed-recurring decimal to fraction:
1. Write down the number consisting with non-repeating digits and repeating digits.
2. Subtract non-repeating number from above.
3. Divide 1-2 by the number with 9's and 0's: for every repeating digit write down a 9, and for every non-repeating digit write down a zero after 9's. | {
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Example #2: Convert $$0.2512(12)$$ to a fraction.
1. The number consisting with non-repeating digits and repeating digits is 2512;
2. Subtract 25 (non-repeating number) from above: 2512-25=2487;
3. Divide 2487 by 9900 (two 9's as there are two digits in 12 and 2 zeros as there are two digits in 25): 2487/9900=829/3300.
Rounding
Rounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4 or smaller, round down (keep the same) the last digit that you keep.
Example:
5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5.
5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5.
5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.
Ratios and Proportions
Given that $$\frac{a}{b}=\frac{c}{d}$$, where a, b, c and d are non-zero real numbers, we can deduce other proportions by simple Algebra. These results are often referred to by the names mentioned along each of the properties obtained.
$$\frac{b}{a}=\frac{d}{c}$$ - invertendo
$$\frac{a}{c}=\frac{b}{d}$$ - alternendo
$$\frac{a+b}{b}=\frac{c+d}{d}$$ - componendo
$$\frac{a-b}{b}=\frac{c-d}{d}$$ - dividendo
$$\frac{a+b}{a-b}=\frac{c+d}{c-d}$$ - componendo & dividendo
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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10 Mar 2010, 05:22
2
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EXPONENTS & ROOTS
This post is a part of [GMAT MATH BOOK]
created by: Bunuel
edited by: bb, walker
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EXPONENTS | {
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--------------------------------------------------------
EXPONENTS
Exponents are a "shortcut" method of showing a number that was multiplied by itself several times. For instance, number $$a$$ multiplied $$n$$ times can be written as $$a^n$$, where $$a$$ represents the base, the number that is multiplied by itself $$n$$ times and $$n$$ represents the exponent. The exponent indicates how many times to multiple the base, $$a$$, by itself.
Exponents one and zero:
$$a^0=1$$ Any nonzero number to the power of 0 is 1.
For example: $$5^0=1$$ and $$(-3)^0=1$$
$$a^1=a$$ Any number to the power 1 is itself.
Powers of zero:
If the exponent is positive, the power of zero is zero: $$0^n = 0$$, where $$n > 0$$.
If the exponent is negative, the power of zero ($$0^n$$, where $$n < 0$$) is undefined, because division by zero is implied.
Powers of one:
$$1^n=1$$ The integer powers of one are one.
Negative powers:
$$a^{-n}=\frac{1}{a^n}$$
Powers of minus one:
If n is an even integer, then $$(-1)^n=1$$.
If n is an odd integer, then $$(-1)^n =-1$$.
Operations involving the same exponents:
Keep the exponent, multiply or divide the bases
$$a^n*b^n=(ab)^n$$
$$\frac{a^n}{b^n}=(\frac{a}{b})^n$$
$$(a^m)^n=a^{mn}$$
$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$
Operations involving the same bases:
Keep the base, add or subtract the exponent (add for multiplication, subtract for division)
$$a^n*a^m=a^{n+m}$$
$$\frac{a^n}{a^m}=a^{n-m}$$
Fraction as power:
$$a^{\frac{1}{n}}=\sqrt[n]{a}$$
$$a^{\frac{m}{n}}=\sqrt[n]{a^m}$$
Exponential Equations:
When solving equations with even exponents, we must consider both positive and negative possibilities for the solutions.
For instance $$a^2=25$$, the two possible solutions are $$5$$ and $$-5$$.
When solving equations with odd exponents, we'll have only one solution.
For instance for $$a^3=8$$, solution is $$a=2$$ and for $$a^3=-8$$, solution is $$a=-2$$. | {
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For instance for $$a^3=8$$, solution is $$a=2$$ and for $$a^3=-8$$, solution is $$a=-2$$.
Exponents and divisibility:
$$a^n-b^n$$ is ALWAYS divisible by $$a-b$$.
$$a^n-b^n$$ is divisible by $$a+b$$ if $$n$$ is even.
$$a^n + b^n$$ is divisible by $$a+b$$ if $$n$$ is odd, and not divisible by a+b if n is even.
Perfect Square
A perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is an perfect square.
There are some tips about the perfect square:
• The number of distinct factors of a perfect square is ALWAYS ODD.
• The sum of distinct factors of a perfect square is ALWAYS ODD.
• A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors.
• Perfect square always has even number of powers of prime factors.
LAST DIGIT OF A PRODUCT
Last $$n$$ digits of a product of integers are last $$n$$ digits of the product of last $$n$$ digits of these integers.
For instance last 2 digits of 845*9512*408*613 would be the last 2 digits of 45*12*8*13=540*104=40*4=160=60
Example: The last digit of 85945*89*58307=5*9*7=45*7=35=5?
LAST DIGIT OF A POWER
Determining the last digit of $$(xyz)^n$$:
1. Last digit of $$(xyz)^n$$ is the same as that of $$z^n$$;
2. Determine the cyclicity number $$c$$ of $$z$$;
3. Find the remainder $$r$$ when $$n$$ divided by the cyclisity;
4. When $$r>0$$, then last digit of $$(xyz)^n$$ is the same as that of $$z^r$$ and when $$r=0$$, then last digit of $$(xyz)^n$$ is the same as that of $$z^c$$, where $$c$$ is the cyclisity number. | {
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• Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.
• Integers ending with 2, 3, 7 and 8 have a cyclicity of 4.
• Integers ending with 4 (eg. $$(xy4)^n$$) have a cyclisity of 2. When n is odd $$(xy4)^n$$ will end with 4 and when n is even $$(xy4)^n$$ will end with 6.
• Integers ending with 9 (eg. $$(xy9)^n$$) have a cyclisity of 2. When n is odd $$(xy9)^n$$ will end with 9 and when n is even $$(xy9)^n$$ will end with 1.
Example: What is the last digit of $$127^{39}$$?
Solution: Last digit of $$127^{39}$$ is the same as that of $$7^{39}$$. Now we should determine the cyclisity of $$7$$:
1. 7^1=7 (last digit is 7)
2. 7^2=9 (last digit is 9)
3. 7^3=3 (last digit is 3)
4. 7^4=1 (last digit is 1)
5. 7^5=7 (last digit is 7 again!)
...
So, the cyclisity of 7 is 4.
Now divide 39 (power) by 4 (cyclisity), remainder is 3.So, the last digit of $$127^{39}$$ is the same as that of the last digit of $$7^{39}$$, is the same as that of the last digit of $$7^3$$, which is $$3$$.
ROOTS
Roots (or radicals) are the "opposite" operation of applying exponents. For instance x^2=16 and square root of 16=4.
General rules:
• $$\sqrt{x}\sqrt{y}=\sqrt{xy}$$ and $$\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}$$.
• $$(\sqrt{x})^n=\sqrt{x^n}$$
• $$x^{\frac{1}{n}}=\sqrt[n]{x}$$
• $$x^{\frac{n}{m}}=\sqrt[m]{x^n}$$
• $${\sqrt{a}}+{\sqrt{b}}\neq{\sqrt{a+b}}$$
• $$\sqrt{x^2}=|x|$$, when $$x\leq{0}$$, then $$\sqrt{x^2}=-x$$ and when $$x\geq{0}$$, then $$\sqrt{x^2}=x$$
• When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root.
That is, $$\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.
• Odd roots will have the same sign as the base of the root. For example, $$\sqrt[3]{125} =5$$ and $$\sqrt[3]{-64} =-4$$. | {
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• For GMAT it's good to memorize following values:
$$\sqrt{2}\approx{1.41}$$
$$\sqrt{3}\approx{1.73}$$
$$\sqrt{5}\approx{2.24}$$
$$\sqrt{6}\approx{2.45}$$
$$\sqrt{7}\approx{2.65}$$
$$\sqrt{8}\approx{2.83}$$
$$\sqrt{10}\approx{3.16}$$
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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10 Mar 2010, 06:48
This is a lucky day for me . Thank you. +1Kudos.
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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02 Apr 2010, 01:42
Great resource. Thanks people!
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20 Jul 2011, 06:52
good stuff. thank you!
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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04 Sep 2012, 20:11
Bunuel wrote:
FRACTIONS
This post is a part of [GMAT MATH BOOK]
created by: Bunuel
edited by: bb, walker
--------------------------------------------------------
The information will be included in future versions of GMAT ToolKit
--------------------------------------------------------
Definition
Fractional numbers are ratios (divisions) of integers. In other words, a fraction is formed by dividing one integer by another integer. Set of Fraction is a subset of the set of Rational Numbers. | {
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Fraction can be expressed in two forms fractional representation $$(\frac{m}{n})$$ and decimal representation $$(a.bcd)$$.
Fractional representation
Fractional representation is a way to express numbers that fall in between integers (note that integers can also be expressed in fractional form). A fraction expresses a part-to-whole relationship in terms of a numerator (the part) and a denominator (the whole).
• The number on top of the fraction is called numerator or nominator. The number on bottom of the fraction is called denominator. In the fraction, $$\frac{9}{7}$$, 9 is the numerator and 7 is denominator.
• Fractions that have a value between 0 and 1 are called proper fraction. The numerator is always smaller than the denominator. $$\frac{1}{3}$$ is a proper fraction.
• Fractions that are greater than 1 are called improper fraction. Improper fraction can also be written as a mixed number. $$\frac{5}{2}$$ is improper fraction.
• An integer combined with a proper fraction is called mixed number. $$4\frac{3}{5}$$ is a mixed number. This can also be written as an improper fraction: $$\frac{23}{5}$$
Converting Improper Fractions
• Converting Improper Fractions to Mixed Fractions:
1. Divide the numerator by the denominator
2. Write down the whole number answer
3. Then write down any remainder above the denominator
Example #1: Convert $$\frac{11}{4}$$ to a mixed fraction.
Solution: Divide $$\frac{11}{4} = 2$$ with a remainder of $$3$$. Write down the $$2$$ and then write down the remainder $$3$$ above the denominator $$4$$, like this: $$2\frac{3}{4}$$ | {
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• Converting Mixed Fractions to Improper Fractions:
1. Multiply the whole number part by the fraction's denominator
2. Add that to the numerator
3. Then write the result on top of the denominator
Example #2: Convert $$3\frac{2}{5}$$ to an improper fraction.
Solution: Multiply the whole number by the denominator: $$3*5=15$$. Add the numerator to that: $$15 + 2 = 17$$. Then write that down above the denominator, like this: $$\frac{17}{5}$$
Reciprocal
Reciprocal for a number $$x$$, denoted by $$\frac{1}{x}$$ or $$x^{-1}$$, is a number which when multiplied by $$x$$ yields $$1$$. The reciprocal of a fraction $$\frac{a}{b}$$ is $$\frac{b}{a}$$. To get the reciprocal of a number, divide 1 by the number. For example reciprocal of $$3$$ is $$\frac{1}{3}$$, reciprocal of $$\frac{5}{6}$$ is $$\frac{6}{5}$$.
Operation on Fractions
To add/subtract fractions with the same denominator, add the numerators and place that sum over the common denominator.
To add/subtract fractions with the different denominator, find the Least Common Denominator (LCD) of the fractions, rename the fractions to have the LCD and add/subtract the numerators of the fractions
Multiplying fractions: To multiply fractions just place the product of the numerators over the product of the denominators.
Dividing fractions: Change the divisor into its reciprocal and then multiply.
Example #1: $$\frac{3}{7}+\frac{2}{3}=\frac{9}{21}+\frac{14}{21}=\frac{23}{21}$$
Example #2: Given $$\frac{\frac{3}{5}}{2}$$, take the reciprocal of $$2$$. The reciprocal is $$\frac{1}{2}$$. Now multiply: $$\frac{3}{5}*\frac{1}{2}=\frac{3}{10}$$.
Decimal Representation
The decimals has ten as its base. Decimals can be terminating (ending) (such as 0.78, 0.2) or repeating (recuring) decimals (such as 0.333333....). | {
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Reduced fraction $$\frac{a}{b}$$ (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only $$b$$ (denominator) is of the form $$2^n5^m$$, where $$m$$ and $$n$$ are non-negative integers. For example: $$\frac{7}{250}$$ is a terminating decimal $$0.028$$, as $$250$$ (denominator) equals to $$2*5^3$$. Fraction $$\frac{3}{30}$$ is also a terminating decimal, as $$\frac{3}{30}=\frac{1}{10}$$ and denominator $$10=2*5$$.
Converting Decimals to Fractions
• To convert a terminating decimal to fraction:
1. Calculate the total numbers after decimal point
2. Remove the decimal point from the number
3. Put 1 under the denominator and annex it with "0" as many as the total in step 1
4. Reduce the fraction to its lowest terms
Example: Convert $$0.56$$ to a fraction.
1: Total number after decimal point is 2.
2 and 3: $$\frac{56}{100}$$.
4: Reducing it to lowest terms: $$\frac{56}{100}=\frac{14}{25}$$
• To convert a recurring decimal to fraction:
1. Separate the recurring number from the decimal fraction
2. Annex denominator with "9" as many times as the length of the recurring number
3. Reduce the fraction to its lowest terms
Example #1: Convert $$0.393939...$$ to a fraction.
1: The recurring number is $$39$$.
2: $$\frac{39}{99}$$, the number $$39$$ is of length $$2$$ so we have added two nines.
3: Reducing it to lowest terms: $$\frac{39}{99}=\frac{13}{33}$$.
• To convert a mixed-recurring decimal to fraction:
1. Write down the number consisting with non-repeating digits and repeating digits.
2. Subtract non-repeating number from above.
3. Divide 1-2 by the number with 9's and 0's: for every repeating digit write down a 9, and for every non-repeating digit write down a zero after 9's. | {
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Example #2: Convert $$0.2512(12)$$ to a fraction.
1. The number consisting with non-repeating digits and repeating digits is 2512;
2. Subtract 25 (non-repeating number) from above: 2512-25=2487;
3. Divide 2487 by 9900 (two 9's as there are two digits in 12 and 2 zeros as there are two digits in 25): 2487/9900=829/3300.
Rounding
Rounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4 or smaller, round down (keep the same) the last digit that you keep.
Example:
5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5.
5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5.
5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.
Ratios and Proportions
Given that $$\frac{a}{b}=\frac{c}{d}$$, where a, b, c and d are non-zero real numbers, we can deduce other proportions by simple Algebra. These results are often referred to by the names mentioned along each of the properties obtained.
$$\frac{b}{a}=\frac{d}{c}$$ - invertendo
$$\frac{a}{c}=\frac{b}{d}$$ - alternendo
$$\frac{a+b}{b}=\frac{c+d}{d}$$ - componendo
$$\frac{a-b}{b}=\frac{c-d}{d}$$ - dividendo
$$\frac{a+b}{a-b}=\frac{c+d}{c-d}$$ - componendo & dividendo
Thanks, this a great and concise summary of all the basics.
Just to add, another important basic rule
for +ve fractions,
if $$\frac{a}{b}<1$$ and x is any positive number then $$\frac{(a+x)}{(b+x)}>\frac{a}{b}$$
,in short when the two fractions have same difference between numerator and denominator , the fraction having greater numerator is always greater.
Also, when $$\frac{a}{b}>1$$ and x is any positive number then $$\frac{(a+x)}{(b+x)}<\frac{a}{b}$$ | {
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Also, when $$\frac{a}{b}>1$$ and x is any positive number then $$\frac{(a+x)}{(b+x)}<\frac{a}{b}$$
Example which one is greater of the following two:
a)$$\frac{111487}{111490}$$ b)$$\frac{111587}{111590}$$ , the ans is b for the same reasons above, as $$\frac{111487+100}{111490+100}$$
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02 Oct 2015, 09:17
• All prime numbers except 2 and 5 end in 1, 3, 7 or 9, since numbers ending in 0, 2, 4, 6 or 8 are multiples of 2 and numbers ending in 0 or 5 are multiples of 5. Similarly, all prime numbers above 3 are of the form 6n−1 or 6n+1, because all other numbers are divisible by 2 or 3.
Why does this property "all prime numbers above 3 are of the form 6n−1 or 6n+1" hold true for 961, but 961 is not a prime? This holds true for square of any prime number? Is there any gap in my understanding?
6n+1 = 961 --> 6n = 960
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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03 Oct 2015, 02:57
Very useful...keep posting..
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03 Oct 2015, 03:43
charan2892 wrote:
Very useful...keep posting..
Thank you.
For more check ALL YOU NEED FOR QUANT ! ! !.
Hope it helps.
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Re: Math: Number Theory (broken into smaller topics) [#permalink]
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06 Dec 2017, 10:05
Bunuel
Thanks so much! Super useful!!
Question: The problem sets referenced at the end of number theory pdf are based on edition 12 of the GMAT official guide. I would like to try prep questions for number theory but don't have this edition of the book (I have the 2017 edition). Where can I find the questions you wrote in the pdf or where can I find questions to test the number theory concepts covered in your pdf?
Many thanks!
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gmatm8 wrote:
Bunuel
Thanks so much! Super useful!!
Question: The problem sets referenced at the end of number theory pdf are based on edition 12 of the GMAT official guide. I would like to try prep questions for number theory but don't have this edition of the book (I have the 2017 edition). Where can I find the questions you wrote in the pdf or where can I find questions to test the number theory concepts covered in your pdf?
Many thanks!
2. Properties of Integers | {
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Many thanks!
2. Properties of Integers
For more check Ultimate GMAT Quantitative Megathread
Hope it helps.
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# How big can a set of integers be if all pairs have small gcd?
Suppose $A\subset[1,N]$ is a set of integers. If for any distinct $a,b\in A$ we have $(a,b)\leq M$ then how big can $|A|$ be?
If $M=1$ then $|A|$ is at most $\pi(N)$ since the map $a\mapsto P_+(a)$ (which sends $a$ to its largest prime factor) is an injection to the primes, and so the primes themselves are the most efficient. If $M=N$ then we can take $|A|=N$. What about $M$ in between? Can you beat the primes? It would be safe to assume that $A$ consists solely of $M$-smooth numbers since we can decompose $A=A_1\cup A_2$ where each $a\in A_1$ is $M$-smooth and each $a\in A_2$ has a prime factor which at least $M+1$. The large prime factors appearing as divisors in $A_2$ cannot be repeated, so $|A_2|\leq \pi(N)$.
• Even more efficiently, you may beat the primes by adding all pairwise products of primes not exceeding $M$ (including their squares). – Ilya Bogdanov Oct 21 '15 at 16:31
• Actually, for $M= 1$ you have $|A| = \pi(N) + 1$: you can have $1$ as well as the primes. @IlyaBogdanov: there's no reason to restrict to pairwise products, you can throw in products of arbitrarily many (not necessarily distinct) primes $x = p_1 \ldots p_n$, $p_1 \le \ldots \le p_n$, as long as $x/p_1 \le M$. – Robert Israel Oct 21 '15 at 18:21
• Thus for $M=5$, in addition to the primes $\le N$ you have $\{1,4,6,8,9,10,15,25\}$. – Robert Israel Oct 21 '15 at 18:28
We'll prove that the maximal cardinality of such a set for $M^2\leq N$ has size equal to $$\pi(N)+\sum_{1<n\leq M} \pi(p(n))$$ where $p(n)$ is the smallest prime factor of $n$. Since $$\sum_{1<n\leq M} \pi(p(n))\sim \frac{M^2}{2\log^2 M},$$ this proves that Ilya Bogdanov's example of including all pairwise products of primes not exceeding $M$ is nearly optimal in terms of asymptotics. | {
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As you suggest in the question, this problem is equivalent to constructing the largest subset $A\subset S_M(N)$ such that $\gcd(a,b)\leq M$ for every $a,b\in A$ where $S_M(N)$ is the set of $n\leq N$ whose largest prime factor is at most $M$.
To see why, suppose that $A$ satisfies $\gcd(a,b)\leq M$ for every $a,b\in A$. Then every prime that is greater than $M$ can divide at most one element of $A$. If $p>M$ divides $a\in A$, then making $a=p$ only helps create a larger set $A$. Since these primes do not interact with the $M$-smooth numbers, the proof is complete.
Let $T(N,M)$ denote the maximum size of such a set $A\subset S_M(N)$. Then the size of the largest subset of $[1,N]$ with pairwise $\gcd$'s bounded by $M$ is $$\pi(N)-\pi(M)+T(N,M).$$
As mentioned by Fedja in the comments s, the reasoning above for the primes extends to all integers. By considering those primes $p,q\leq M$ such that $pq>M$, we see that there can only be exactly one such number divisible by $pq$. Similarly for any $p,q,r$ with $pq,qr,rp\leq M$ and $pqr>M$ there can only be one number in our set divisible by $pqr$. Thus we find that the maximal size is the sum over those integers whose largest proper divisor is less than $M$. Since the largest proper divisor of $n$ equals $n/p(n)$ where $p(n)$ is the least prime factor, we can group things based on this, and we have that $$T(N,M)=\sum_{1<p\leq M}\sum_{n\leq M:\ p(n)\geq p}1.$$ Rearranging this equals $$\sum_{1<n\leq M}\sum_{p\leq p(n)}1=\sum_{n\leq M}\pi\left(p(n)\right),$$ and for composite $n$ $p(n)\leq\sqrt{n}$, so the primes dominate this sum. Thus asymptotically we have $$T(N,M)\sim\sum_{p\leq M}\pi(p)\sim\frac{M^{2}}{2\log^{2}M}.$$ | {
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• You can continue in the same spirit: consider pairs of primes $p,q\le M$ with $pq>M$. Any such pair can enter just one number, so we can as well have all pairs there. Next consider all triples such that the product of any 2 is less than or equal to $M$ but the triple product is $>M$. Again, any triple can enter at most one number, so take them all, etc. After that, of course, just add $[1,M]$. So to describe a maximal cardinality set it easy. As to the size, it looks like we just need to play with the prime number theorem carefully to figure the asymptotics but I have to teach a class now... – fedja Oct 21 '15 at 18:08
• This set can be also described as the set of all numbers whose largest proper divisor is not greater than $M$, which allows to estimate the size of the complement way faster than with my prime count idea. Now it is time to run on the stairs :-) – fedja Oct 21 '15 at 18:13
• @Fedja: Great, that works nicely! – Eric Naslund Oct 21 '15 at 18:22
• There's still something off if $M>\sqrt{N}$ which you should account for. Fedja's description of numbers with largest proper divisor not larger than $M$ is very nice. For example, this includes all numbers with no prime factors below $N/M$ which has interesting asymptotics when $M$ is larger than $\sqrt{N}$ (eventually getting to $N$ when $M=N$). So you could either restrict attention to small ranges of $M$, or perhaps flesh out the argument a little bit more ... – Lucia Oct 22 '15 at 3:19
• @Lucia: You're right, I have edited the question for now, but I'll add in a nice form for the sum when $M>\sqrt{N}$ – Eric Naslund Oct 22 '15 at 11:09 | {
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03 413 022
### Question 1
Up to now our machines have consisted of a row of boxes extending infinitely far to the left. Why not have boxes extending infinitely far to the right as well?
Let’s go back to working with a $1 \leftarrow 10$ machine and see what such boxes could mean in that machine.
(Then it should become clear what they mean in other machines as well.)
To keep the left and right boxes visibly clear, we’ll separate them with a point. (Society calls this point, for a $1 \leftarrow 10$ machine at least, a decimal point.)
So, what does it mean to have dots in the right boxes? What are the values of dots in those boxes?
Since this is a $1 \leftarrow 10$ machine, we do know that ten dots in any one box explode to make one dot one place to the left. So ten dots in the box just to the right of the decimal point are equivalent to one dot in the $1$ s box. Each dot in that box must be worth one-tenth.
We have:
In the same way, ten dots in the next box over are worth one-tenth. And so each dot in that next box must be worth one-hundredth.
We have:
And ten one-thousands make a hundredth, and ten ten-thousands make a thousandth, and so on.
We see that the boxes to the left of the decimal point represent place values as given by the powers of ten, and the boxes to the right of the decimal point place values given by the reciprocals of the powers of ten.
We have just discovered decimals!
When people write $0.3$, for example, they mean the value of placing three dots in the first box after the decimal point.
We see that $0.3$ equals three tenths: $0.3 = \dfrac{3}{10}$.
Seven dots in the third box after the decimal point is seven thousandths: $0.007 = \dfrac{7}{1000}$.
Comment: Some people might leave off the beginning zero and just write $.007 = \dfrac{7}{1000}$. It’s just a matter of personal taste. | {
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Some people read $0.6$ out loud as “point six” and others read it out loud as “six tenths.” Which is more helpful for understanding what the number really is?
### Question 2
There is a possible source of confusion with a decimal such as $0.31$. This is technically three tenths and one hundredth: $0.31 = \dfrac{3}{10} + \dfrac{1}{100}$.
But some people read $0.31$ out loud as “thirty-one hundredths,” which looks like this.
Are these the same thing?
Well, yes! With three explosions we see that thirty-one hundredths becomes three tenths and one hundredth.
Comment: You can also show that $\dfrac{3}{10} + \dfrac{1}{100}$ and $\dfrac{31}{100}$ are the same with the arithmetic of adding fractions. We have
$\dfrac{3}{10} + \dfrac{1}{100} = \dfrac{30}{100} + \dfrac{1}{100} = \dfrac{31}{100}$.
(Do you see that this is really the result of performing three unexplosions in a picture of $\dfrac{3}{10} + \dfrac{1}{100}$?)
A teacher asked his students to each draw a $1 \leftarrow 10$ machine picture of the fraction $\dfrac{319}{1000}$.
JinJin drew:
Subra drew:
The teacher marked both students as correct. Are each of these solutions indeed valid? Explain your thinking. (By the way, the teacher doesn’t mind if students just write numbers instead of drawing dots.)
### Question 3
Multiple choice!
The decimal $0.23$ equals:
(A) $\dfrac {23}{10}$
(B) $\dfrac {23}{100}$
(C) $\dfrac {23}{1000}$
(D) $\dfrac {23}{10000}$
### Question 4
The decimal $0.0409$ equals:
(A) $\dfrac {409}{100}$
(B) $\dfrac {409}{1000}$
(C) $\dfrac {409}{10000}$
(D) $\dfrac {409}{100000}$
## SIMPLE FRACTIONS AS DECIMALS
Some decimals give fractions that simplify further.
For example,
$0.5 = \dfrac{5}{10} = \dfrac{1}{2}$
and
$0.04 = \dfrac{4}{100} = \dfrac{1}{25}$.
Conversely, if a fraction can be rewritten to have a denominator that is a power of ten, then we can easily write it as a decimal.
For example,
$\dfrac{3}{5} = \dfrac{6}{10}$ and so $\dfrac{3}{5} = 0.6$
and | {
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For example,
$\dfrac{3}{5} = \dfrac{6}{10}$ and so $\dfrac{3}{5} = 0.6$
and
$\dfrac{13}{20} = \dfrac{13 \times 5}{20 \times 5} = \dfrac{65}{100} = 0.65$.
What fractions (in simplest terms) do the following decimals represent?
$0.05$, $0.2$, $0.8$, $0.004$
### Question 6
Write each of the following fractions as a decimal.
$\dfrac {2} {5}$, $\dfrac {1} {25}$, $\dfrac {1} {20}$, $\dfrac {1} {200}$, $\dfrac {2} {2500}$
### Question 7
MULTIPLE CHOICE!
The decimal $0.050$ equals
(A)$\dfrac {50} {100}$
(B)$\dfrac {1} {20}$
(C)$\dfrac {1} {200}$
(D) None of these?
### Question 8
The decimal $0.000208$ equals
(A) $\dfrac {52} {250}$
(B) $\dfrac {52} {2500}$
(C) $\dfrac {52} {25000}$
(D) $\dfrac {52} {250000}$
### Question 9
Write each of the following fractions as decimals.
$\dfrac {7} {20}$, $\dfrac {16} {25}$, $\dfrac {301} {500}$, $\dfrac {17} {50}$, $\dfrac {3} {4}$
### Question 10
CHALLENGE
What fraction does the decimal $2.3$ represent?
### Question 11
What fraction does $17.04$ represent?
### Question 12
What fraction does $1003.1003$ represent?
### Question 13
Let’s explore the question: Do $0.19$ and $0.190$ represent the same number or different numbers?
Here are two dots and boxes pictures for the decimal $0.19$:
Here are two dots and boxes picture for the decimal $0.190$
(A) Explain how one “unexplosion” establishes that the first picture of $0.19$ is equivalent to the second picture of $0.19$.
(B) Explain how several unexplosions establishes that the first picture of $0.190$ is equivalent to the second picture of $0.190$.
(C) Explain how explosions and unexplosions in fact establish that all four pictures are equivalent to each other.
(D) In conclusion then: Does $0.190$ represent the same number as $0.19$?
### Question 14
To a mathematician, the expressions $0.19$ and $0.190$ represent exactly the same numeric quantity. | {
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To a mathematician, the expressions $0.19$ and $0.190$ represent exactly the same numeric quantity.
But you may have noticed in science class that scientists will often write down what seems likes
unnecessary zeros when recording measurements. This is because scientists want to impart more information to the reader than just a numeric value.
For example, suppose a botanist measures the length of a stalk. By writing the measurement as $0.190$ meters in her paper, the scientist is saying to the reader that she measured the length of the stalk to the nearest one thousandth of a meter and that she got $1$ tenth, $9$ hundredths, and $0$ thousandths of a meter. Thus we are being told that the true length of the stalk is somewhere in the range of $0.1895$ and $0.1905$ meters.
If she wrote in her paper, instead, just $0.19$ meters, then we would have to assume she measured the length of the stalk only to the nearest hundredth of a meter and so its true length lies somewhere between $0.185$ and $0.195$ meters.
## MIXED NUMBERS AS DECIMALS.
How does $12 \dfrac {3} {4}$, for example, appear as a decimal?
Well, $12 \dfrac {3} {4} = 12 + \dfrac {3} {4}$ and we can certainly write the fractional part as a decimal. (The non-fractional part is already in the $1 \leftarrow 10$ machine format!)
We have
$12 \dfrac {3} {4} = 12 + \dfrac {75} {100}$
and so we see
$12 \dfrac {3} {4} = 12.75$.
Write each of the following numbers in decimal notation.
(A) $5 \dfrac {3} {10}$
(B) $7 \dfrac {1} {5}$
(C) $13 \dfrac {1} {2}$
(D) $106 \dfrac {3} {20}$
(E) $\dfrac {78} {25}$
(F) $\dfrac {9} {4}$
(G) $\dfrac {131} {40}$
You can either play with some of the optional stations below or go to the next island!
Register NOW and unlock all islands! | {
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# Solving $\sin(\omega t)=- \frac 1 2$
Solve the given trigonometric equation: $$\sin(\omega t)=- \frac 1 2$$
Here is my attempt:
$$\sin(\omega t)=- \frac 1 2 = \sin \biggr (\pi +\dfrac{\pi}{6}\biggr )$$
Which yields
$$\omega t = \dfrac{7\pi }{6}$$
or
$$\sin(\omega t)=- \frac 1 2 = \sin \biggr (2\pi -\dfrac{\pi}{6}\biggr )$$
$$\omega t = \dfrac{11\pi}{6}$$
Is my assumption correct?
Regards!
• I've fixed a small error. – Mr. Maxwell Jul 7 '18 at 21:17
You also have $$\omega t = \dfrac{-\pi }{6}+2k\pi$$
Thus the solution is $$\omega t = \{\dfrac{-\pi }{6}+2k_1\pi:k_1 \in Z\}\cup \{\dfrac{7\pi }{6}+2k_2\pi:k_2\in Z\}$$
Where Z is the set of integers.
From the unit circle definition of sine, we see that
$$\omega t = 2n\pi - \dfrac{\pi}{2} \pm \dfrac{\pi}{3}$$
and we can rephrase that as we wish.
The sin$(x)$ function repeats itself every $2\pi$ times, so $\omega t=\frac{7\pi}6+2kπ$ and $\omega t=\frac{11\pi}{6}+2k\pi,\;$ where $k \in \mathbb Z$.
Yes it is right but recall to add the $2k\pi$ term with $k\in \mathbb{Z}$, indeed we have
$$\sin(\omega t)=- \frac 1 2\iff \omega t=\dfrac{7\pi }{6}+2k\pi\,\lor\,\omega t=\dfrac{11\pi }{6}+2k\pi$$
• What if I don't recall to add $2k\pi$? – Mr. Maxwell Jul 7 '18 at 21:16
• @Mr.Maxwell You can lost solutions. – user Jul 7 '18 at 21:17
• However, isn't it enough to leave it up once we get $\dfrac{7\pi}{6}$ and $\dfrac{11\pi}{6}$? – Mr. Maxwell Jul 7 '18 at 21:18
• @Mr.Maxwell Those are the solutions for $\omega t \in [0,2\pi]$ but in general solving trigonometric equations we need to consider a wider range. Take a look for example here math.stackexchange.com/q/2826222/505767 – user Jul 7 '18 at 21:20
• @Mr.Maxwell - Enough is defined by who is asking the question. If the domain is not specified, then the question is poorly asked. – steven gregory Jul 7 '18 at 22:31 | {
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plot3
3-D point or line plot
Description
example
plot3(X,Y,Z) plots coordinates in 3-D space.
• To plot a set of coordinates connected by line segments, specify X, Y, and Z as vectors of the same length.
• To plot multiple sets of coordinates on the same set of axes, specify at least one of X, Y, or Z as a matrix and the others as vectors.
example
plot3(X,Y,Z,LineSpec) creates the plot using the specified line style, marker, and color.
example
plot3(X1,Y1,Z1,...,Xn,Yn,Zn) plots multiple sets of coordinates on the same set of axes. Use this syntax as an alternative to specifying multiple sets as matrices.
example
plot3(X1,Y1,Z1,LineSpec1,...,Xn,Yn,Zn,LineSpecn) assigns specific line styles, markers, and colors to each XYZ triplet. You can specify LineSpec for some triplets and omit it for others. For example, plot3(X1,Y1,Z1,'o',X2,Y2,Z2) specifies markers for the first triplet but not the for the second triplet.
example
plot3(___,Name,Value) specifies Line properties using one or more name-value pair arguments. Specify the properties after all other input arguments. For a list of properties, see Line Properties.
example
plot3(ax,___) displays the plot in the target axes. Specify the axes as the first argument in any of the previous syntaxes.
example
p = plot3(___) returns a Line object or an array of Line objects. Use p to modify properties of the plot after creating it. For a list of properties, see Line Properties.
Examples
collapse all
Define t as a vector of values between 0 and 10$\pi$. Define st and ct as vectors of sine and cosine values. Then plot st, ct, and t.
t = 0:pi/50:10*pi;
st = sin(t);
ct = cos(t);
plot3(st,ct,t)
Create two sets of x-, y-, and z-coordinates.
t = 0:pi/500:pi;
xt1 = sin(t).*cos(10*t);
yt1 = sin(t).*sin(10*t);
zt1 = cos(t);
xt2 = sin(t).*cos(12*t);
yt2 = sin(t).*sin(12*t);
zt2 = cos(t);
Call the plot3 function, and specify consecutive XYZ triplets.
plot3(xt1,yt1,zt1,xt2,yt2,zt2) | {
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"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982287698703999,
"lm_q1q2_score": 0.8667980397304585,
"lm_q2_score": 0.8824278680004707,
"openwebmath_perplexity": 2821.3270388311807,
"openwebmath_score": 0.5981382131576538,
"tags": null,
"url": "https://au.mathworks.com/help/matlab/ref/plot3.html"
} |
Call the plot3 function, and specify consecutive XYZ triplets.
plot3(xt1,yt1,zt1,xt2,yt2,zt2)
Create matrix X containing three rows of x-coordinates. Create matrix Y containing three rows of y-coordinates.
t = 0:pi/500:pi;
X(1,:) = sin(t).*cos(10*t);
X(2,:) = sin(t).*cos(12*t);
X(3,:) = sin(t).*cos(20*t);
Y(1,:) = sin(t).*sin(10*t);
Y(2,:) = sin(t).*sin(12*t);
Y(3,:) = sin(t).*sin(20*t);
Create matrix Z containing the z-coordinates for all three sets.
Z = cos(t);
Plot all three sets of coordinates on the same set of axes.
plot3(X,Y,Z)
Create vectors xt, yt, and zt.
t = 0:pi/500:40*pi;
xt = (3 + cos(sqrt(32)*t)).*cos(t);
yt = sin(sqrt(32) * t);
zt = (3 + cos(sqrt(32)*t)).*sin(t);
Plot the data, and use the axis equal command to space the tick units equally along each axis. Then specify the labels for each axis.
plot3(xt,yt,zt)
axis equal
xlabel('x(t)')
ylabel('y(t)')
zlabel('z(t)')
Create vectors t, xt, and yt, and plot the points in those vectors using circular markers.
t = 0:pi/20:10*pi;
xt = sin(t);
yt = cos(t);
plot3(xt,yt,t,'o')
Create vectors t, xt, and yt, and plot the points in those vectors as a blue line with 10-point circular markers. Use a hexadecimal color code to specify a light blue fill color for the markers.
t = 0:pi/20:10*pi;
xt = sin(t);
yt = cos(t);
plot3(xt,yt,t,'-o','Color','b','MarkerSize',10,'MarkerFaceColor','#D9FFFF')
Create vector t. Then use t to calculate two sets of x and y values.
t = 0:pi/20:10*pi;
xt1 = sin(t);
yt1 = cos(t);
xt2 = sin(2*t);
yt2 = cos(2*t);
Plot the two sets of values. Use the default line for the first set, and specify a dashed line for the second set.
plot3(xt1,yt1,t,xt2,yt2,t,'--')
Create vectors t, xt, and yt, and plot the data in those vectors. Return the chart line in the output variable p.
t = linspace(-10,10,1000);
xt = exp(-t./10).*sin(5*t);
yt = exp(-t./10).*cos(5*t);
p = plot3(xt,yt,t);
Change the line width to 3.
p.LineWidth = 3; | {
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"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982287698703999,
"lm_q1q2_score": 0.8667980397304585,
"lm_q2_score": 0.8824278680004707,
"openwebmath_perplexity": 2821.3270388311807,
"openwebmath_score": 0.5981382131576538,
"tags": null,
"url": "https://au.mathworks.com/help/matlab/ref/plot3.html"
} |
Change the line width to 3.
p.LineWidth = 3;
Starting in R2019b, you can display a tiling of plots using the tiledlayout and nexttile functions. Call the tiledlayout function to create a 1-by-2 tiled chart layout. Call the nexttile function to create the axes objects ax1 and ax2. Create separate line plots in the axes by specifying the axes object as the first argument to plot3.
tiledlayout(1,2)
% Left plot
ax1 = nexttile;
t = 0:pi/20:10*pi;
xt1 = sin(t);
yt1 = cos(t);
plot3(ax1,xt1,yt1,t)
title(ax1,'Helix With 5 Turns')
% Right plot
ax2 = nexttile;
t = 0:pi/20:10*pi;
xt2 = sin(2*t);
yt2 = cos(2*t);
plot3(ax2,xt2,yt2,t)
title(ax2,'Helix With 10 Turns')
Create x and y as vectors of random values between 0 and 1. Create z as a vector of random duration values.
x = rand(1,10);
y = rand(1,10);
z = duration(rand(10,1),randi(60,10,1),randi(60,10,1));
Plot x, y, and z, and specify the format for the z-axis as minutes and seconds. Then add axis labels, and turn on the grid to make it easier to visualize the points within the plot box.
plot3(x,y,z,'o','DurationTickFormat','mm:ss')
xlabel('X')
ylabel('Y')
zlabel('Duration')
grid on
Create vectors xt, yt, and zt. Plot the values, specifying a solid line with circular markers using the LineSpec argument. Specify the MarkerIndices property to place one marker at the 200th data point.
t = 0:pi/500:pi;
xt(1,:) = sin(t).*cos(10*t);
yt(1,:) = sin(t).*sin(10*t);
zt = cos(t);
plot3(xt,yt,zt,'-o','MarkerIndices',200)
Input Arguments
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x-coordinates, specified as a scalar, vector, or matrix. The size and shape of X depends on the shape of your data and the type of plot you want to create. This table describes the most common situations.
Type of PlotHow to Specify Coordinates
Single point
Specify X, Y, and Z as scalars and include a marker. For example:
plot3(1,2,3,'o')
One set of points
Specify X, Y, and Z as any combination of row or column vectors of the same length. For example: | {
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"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982287698703999,
"lm_q1q2_score": 0.8667980397304585,
"lm_q2_score": 0.8824278680004707,
"openwebmath_perplexity": 2821.3270388311807,
"openwebmath_score": 0.5981382131576538,
"tags": null,
"url": "https://au.mathworks.com/help/matlab/ref/plot3.html"
} |
Specify X, Y, and Z as any combination of row or column vectors of the same length. For example:
plot3([1 2 3],[4; 5; 6],[7 8 9])
Multiple sets of points
(using vectors)
Specify consecutive sets of X, Y, and Z vectors. For example:
plot3([1 2 3],[4 5 6],[7 8 9],[1 2 3],[4 5 6],[10 11 12])
Multiple sets of points
(using matrices)
Specify at least one of X, Y, or Z as a matrix, and the others as vectors. Each of X, Y, and Z must have at least one dimension that is same size. For best results, specify all vectors of the same shape and all matrices of the same shape. For example:
plot3([1 2 3],[4 5 6],[7 8 9; 10 11 12])
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64 | categorical | datetime | duration
y-coordinates, specified as a scalar, vector, or matrix. The size and shape of Y depends on the shape of your data and the type of plot you want to create. This table describes the most common situations.
Type of PlotHow to Specify Coordinates
Single point
Specify X, Y, and Z as scalars and include a marker. For example:
plot3(1,2,3,'o')
One set of points
Specify X, Y, and Z as any combination of row or column vectors of the same length. For example:
plot3([1 2 3],[4; 5; 6],[7 8 9])
Multiple sets of points
(using vectors)
Specify consecutive sets of X, Y, and Z vectors. For example:
plot3([1 2 3],[4 5 6],[7 8 9],[1 2 3],[4 5 6],[10 11 12])
Multiple sets of points
(using matrices)
Specify at least one of X, Y, or Z as a matrix, and the others as vectors. Each of X, Y, and Z must have at least one dimension that is same size. For best results, specify all vectors of the same shape and all matrices of the same shape. For example:
plot3([1 2 3],[4 5 6],[7 8 9; 10 11 12])
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64 | categorical | datetime | duration | {
"domain": "mathworks.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982287698703999,
"lm_q1q2_score": 0.8667980397304585,
"lm_q2_score": 0.8824278680004707,
"openwebmath_perplexity": 2821.3270388311807,
"openwebmath_score": 0.5981382131576538,
"tags": null,
"url": "https://au.mathworks.com/help/matlab/ref/plot3.html"
} |
z-coordinates, specified as a scalar, vector, or matrix. The size and shape of Z depends on the shape of your data and the type of plot you want to create. This table describes the most common situations.
Type of PlotHow to Specify Coordinates
Single point
Specify X, Y, and Z as scalars and include a marker. For example:
plot3(1,2,3,'o')
One set of points
Specify X, Y, and Z as any combination of row or column vectors of the same length. For example:
plot3([1 2 3],[4; 5; 6],[7 8 9])
Multiple sets of points
(using vectors)
Specify consecutive sets of X, Y, and Z vectors. For example:
plot3([1 2 3],[4 5 6],[7 8 9],[1 2 3],[4 5 6],[10 11 12])
Multiple sets of points
(using matrices)
Specify at least one of X, Y, or Z as a matrix, and the others as vectors. Each of X, Y, and Z must have at least one dimension that is same size. For best results, specify all vectors of the same shape and all matrices of the same shape. For example:
plot3([1 2 3],[4 5 6],[7 8 9; 10 11 12])
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64 | categorical | datetime | duration
Line style, marker, and color, specified as a character vector or string containing symbols. The symbols can appear in any order. You do not need to specify all three characteristics (line style, marker, and color). For example, if you omit the line style and specify the marker, then the plot shows only the marker and no line.
Example: '--or' is a red dashed line with circle markers
Line StyleDescription
-Solid line (default)
--Dashed line
:Dotted line
-.Dash-dot line
MarkerDescription
oCircle
+Plus sign
*Asterisk
.Point
xCross
sSquare
dDiamond
^Upward-pointing triangle
vDownward-pointing triangle
>Right-pointing triangle
<Left-pointing triangle
pPentagram
hHexagram
ColorDescription
y
yellow
m
magenta
c
cyan
r
red
g
green
b
blue
w
white
k
black | {
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"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982287698703999,
"lm_q1q2_score": 0.8667980397304585,
"lm_q2_score": 0.8824278680004707,
"openwebmath_perplexity": 2821.3270388311807,
"openwebmath_score": 0.5981382131576538,
"tags": null,
"url": "https://au.mathworks.com/help/matlab/ref/plot3.html"
} |
y
yellow
m
magenta
c
cyan
r
red
g
green
b
blue
w
white
k
black
Target axes, specified as an Axes object. If you do not specify the axes and if the current axes is Cartesian, then plot3 uses the current axes.
Name-Value Pair Arguments
Specify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.
Example: plot3([1 2],[3 4],[5 6],'Color','red') specifies a red line for the plot.
Note
The properties listed here are only a subset. For a complete list, see Line Properties.
Color, specified as an RGB triplet, a hexadecimal color code, a color name, or a short name. The color you specify sets the line color. It also sets the marker edge color when the MarkerEdgeColor property is set to 'auto'.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color | {
"domain": "mathworks.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982287698703999,
"lm_q1q2_score": 0.8667980397304585,
"lm_q2_score": 0.8824278680004707,
"openwebmath_perplexity": 2821.3270388311807,
"openwebmath_score": 0.5981382131576538,
"tags": null,
"url": "https://au.mathworks.com/help/matlab/ref/plot3.html"
} |
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB® uses in many types of plots.
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Line width, specified as a positive value in points, where 1 point = 1/72 of an inch. If the line has markers, then the line width also affects the marker edges.
The line width cannot be thinner than the width of a pixel. If you set the line width to a value that is less than the width of a pixel on your system, the line displays as one pixel wide.
Marker size, specified as a positive value in points, where 1 point = 1/72 of an inch.
Marker outline color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The default value of 'auto' uses the same color as the Color property.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF' | {
"domain": "mathworks.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982287698703999,
"lm_q1q2_score": 0.8667980397304585,
"lm_q2_score": 0.8824278680004707,
"openwebmath_perplexity": 2821.3270388311807,
"openwebmath_score": 0.5981382131576538,
"tags": null,
"url": "https://au.mathworks.com/help/matlab/ref/plot3.html"
} |
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Marker fill color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The 'auto' option uses the same color as the Color property of the parent axes. If you specify 'auto' and the axes plot box is invisible, the marker fill color is the color of the figure.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color | {
"domain": "mathworks.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982287698703999,
"lm_q1q2_score": 0.8667980397304585,
"lm_q2_score": 0.8824278680004707,
"openwebmath_perplexity": 2821.3270388311807,
"openwebmath_score": 0.5981382131576538,
"tags": null,
"url": "https://au.mathworks.com/help/matlab/ref/plot3.html"
} |
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Tips
• Use NaN or Inf to create breaks in the lines. For example, this code plots a line with a break between z=2 and z=4.
plot3([1 2 3 4 5],[1 2 3 4 5],[1 2 NaN 4 5])
• plot3 uses colors and line styles based on the ColorOrder and LineStyleOrder properties of the axes. plot3 cycles through the colors with the first line style. Then, it cycles through the colors again with each additional line style.
Starting in R2019b, you can change the colors and the line styles after plotting by setting the ColorOrder or LineStyleOrder properties on the axes. You can also call the colororder function to change the color order for all the axes in the figure. | {
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"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.982287698703999,
"lm_q1q2_score": 0.8667980397304585,
"lm_q2_score": 0.8824278680004707,
"openwebmath_perplexity": 2821.3270388311807,
"openwebmath_score": 0.5981382131576538,
"tags": null,
"url": "https://au.mathworks.com/help/matlab/ref/plot3.html"
} |
# Math Help - Integrate LN2x/x^2
1. ## Integrate LN2x/x^2
Hey guys, thanks for all the help so far.. stuck on another problem.
$\int \frac{ln2x}{x^2}$
So I think
$ln2x$
is an integration by parts problem by itself, but I am lost.
I know that
$\int lnx = xlnx-x$
But that was just a rule given by the professor, not sure how we applied that.
I think it goes something like this (using integration by parts)
$u= ln2x \ du= \frac{2x}{x} \ dv=dx \ v=x$
Using the rule
$UV- \int VdU$
I get
$(xln2x)- \int x \frac{2x}{x}$
which equals
$(xln2x)- \int \frac{2x^2}{x}$
But that's just for the top portion of the original and I am not even sure I did that correctly.
Thanks guys
2. Just in case a picture helps...
... where (key in first spoiler) ...
Spoiler:
... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x. And,
... is lazy integration by parts, doing without u and v.
Spoiler:
________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
3. Originally Posted by Spoolx
Hey guys, thanks for all the help so far.. stuck on another problem.
$\int \frac{ln2x}{x^2}$
So I think
$ln2x$
is an integration by parts problem by itself, but I am lost.
I know that
$\int lnx = xlnx-x$
But that was just a rule given by the professor, not sure how we applied that.
I think it goes something like this (using integration by parts)
$u= ln2x \ du= \frac{2x}{x} \ dv=dx \ v=x$
Using the rule
$UV- \int VdU$
I get
$(xln2x)- \int x \frac{2x}{x}$
which equals
$(xln2x)- \int \frac{2x^2}{x}$
But that's just for the top portion of the original and I am not even sure I did that correctly.
Thanks guys
$ln(2x)=ln2+lnx$
$\displaystyle\int{\frac{ln(2x)}{x^2}}dx=ln2\int{\f rac{1}{x^2}}dx+\int{\frac{lnx}{x^2}}dx$
Now you can use integration by parts.
4. Hello, Spoolx!
$\displaystyle \int \frac{\ln(2x)}{x^2}\,dx$ | {
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"url": "http://mathhelpforum.com/calculus/167221-integrate-ln2x-x-2-a.html"
} |
4. Hello, Spoolx!
$\displaystyle \int \frac{\ln(2x)}{x^2}\,dx$
Integrate by parts:
. . $\begin{array}{cccccccc}
u &=& \ln(2x) && dv &=& \dfrac{1}{x^2}\,dx \\ \\[-3mm]
du &=& \dfrac{dx}{x} && v &=& \text{-}\dfrac{1}{x} \end{array}$
And we have: . $\displaystyle \text{-}\frac{1}{x}\ln(2x) + \int\frac{dx}{x^2}$
. . . . . . . . $\displaystyle =\;\text{-}\frac{1}{x}\ln(2x) - \frac{1}{x} + C$
. . . . . . . . $\displaystyle =\;\text{-}\frac{1}{x}\left[\ln(2x) + 1\right] + C$
5. Thanks for the help guys, can you jst run through the derivation of the $ln2x$ please?
6. Just in case a picture helps...
... where (key in spoiler) ...
Spoiler:
... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case ), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).
The general drift is...
Though, as Archie Meade points out, you can split the log first to avoid the chain rule.
_________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
7. Originally Posted by Spoolx
Hey guys, thanks for all the help so far.. stuck on another problem.
$\int \frac{ln2x}{x^2}$
So I think
$ln2x$
is an integration by parts problem by itself, but I am lost.
I know that
$\int lnx = xlnx-x$
But that was just a rule given by the professor, not sure how we applied that.
I think it goes something like this (using integration by parts)
$u= ln2x \ du= \frac{2x}{x} \ dv=dx \ v=x$
Using the rule
$UV- \int VdU$
I get
$(xln2x)- \int x \frac{2x}{x}$
which equals
$(xln2x)- \int \frac{2x^2}{x}$
But that's just for the top portion of the original and I am not even sure I did that correctly. | {
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But that's just for the top portion of the original and I am not even sure I did that correctly.
Thanks guys
Rather than trying to apply integration by parts just to the numerator,
you should try applying the technique to the entire expression.
$\displaystyle\int{u}dv=uv-\int{v}du$
$u=ln(2x),\;\;dv=\displaystyle\frac{1}{x^2}dx$
We need $v$ and $du$ to apply integration by parts. Using the chain rule we get
$\displaystyle\frac{du}{dx}=\frac{1}{2x}2$
$\Rightarrow\ du=\frac{1}{x}dx$
$v=\int{x^{-2}}dx=-x^{-1}$
$\displaystyle\ uv-\int{v}du}=-\frac{ln(2x)}{x}+\int{\frac{1}{x^2}dx=-\frac{ln(2x)}{x}-\frac{1}{x}+C$
Alternatively, begin with
$ln(2x)=ln2+lnx$
8. Glad I found this forum, learning alot from you guys!
Thanks! | {
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# Math Help - verifying answers for polynomials
1. ## verifying answers for polynomials
I would like to verify that the answers that I have come up with are correct on the following 3 equations. I greatly appreciate it! (the numbers in red are exponents) 1) solve the equation 1/4a2 + 3/4 a=1 My answer is a=4 and a=1 ??? 2) Simplify the expression 2ab4 - 3 a2b2 - ab4 + a2b2 I have ab4 - 2a2b2 ??? and 3) Simplify the expression -2(4y2+3z3+5) + 3(2y2-5z3+3) My answer is -21z3 - 2y2 -1??? I thank you for the help!
2. ## latex
Originally Posted by jay1
I would like to verify that the answers that I have come up with are correct on the following 3 equations. I greatly appreciate it! (the numbers in red are exponents) 1) solve the equation 1/4a2 + 3/4 a=1 My answer is a=4 and a=1 ??? 2) Simplify the expression 2ab4 - 3 a2b2 - ab4 + a2b2 I have ab4 - 2a2b2 ??? and 3) Simplify the expression -2(4y2+3z3+5) + 3(2y2-5z3+3) My answer is -21z3 - 2y2 -1??? I thank you for the help!
hope you learn to use laTex
this is my understanding of your equation
$
\frac{1}{4a^2} + \frac{3}{4a} = 1
$
$\frac{1}{4a^2} + \frac{3}{4a}\times\frac{a}{a} \Rightarrow \frac{1+3a}{4a^2} = 1$
$4a^2 - 3a -1 = 0$
$(4a-1)(a+1)= 0$
$a= -1$
$a=-\frac{1}{4}$
3. Is there a place that will show me how to use this "laTex"? BigWave, was my answer correct? Thanks for your help.
4. Originally Posted by jay1
I would like to verify that the answers that I have come up with are correct on the following 3 equations. I greatly appreciate it! (the numbers in red are exponents) 1) solve the equation 1/4a2 + 3/4 a=1 My answer is a=4 and a=1 ??? 2) Simplify the expression 2ab4 - 3 a2b2 - ab4 + a2b2 I have ab4 - 2a2b2 ??? and 3) Simplify the expression -2(4y2+3z3+5) + 3(2y2-5z3+3) My answer is -21z3 - 2y2 -1??? I thank you for the help!
If problem 1 is what I interpreted it to be, which is this:
$
\frac{1}{4a^2} + \frac{3}{4}a = 1
$
then
$
a=1, a=\frac{1+\sqrt{13}}{6}, \text{ and } a=\frac{1-\sqrt{13}}{6}
$
. | {
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then
$
a=1, a=\frac{1+\sqrt{13}}{6}, \text{ and } a=\frac{1-\sqrt{13}}{6}
$
.
However, if problem 1 is this:
$
\frac{1}{4a^2} + \frac{3}{4a} = 1
$
then
$
a=1, a= -\frac{1}{4}
$
.
And if problem 1 is this:
$
\frac{1}{4}a^2 + \frac{3}{4}a = 1
$
then
$
a=1, a= -4
$
.
5. Hello, jay1!
1) Solve: . $\tfrac{1}{4}a^2 + \tfrac{3}{4}a \:=\:1$
My answer: . $a=4,\;a=1$ . . . . no
We have: . $\tfrac{1}{4}a^2 + \tfrac{3}{4}a \:=\:1$
Multiply by 4: . $a^2 + 3a \:=\:4 \quad\Rightarrow\quad a^2 + 3a - 4 \:=\:0$
Factor: . $(a-1)(a+4) \:=\:0$
Therefore: . $a\;=\;1,\:-4$
2) Simplify: . $2ab^4 -3a^2b^2 - ab^4 + a^2b^2$
I have: . $ab^4- 2a^2b^2$ . . . . Yes!
3) Simplify: . $-2(4y^2 +3z^3 +5) + 3(2y^2 -5z^3 +3)$
My answer: . $-21z^3 - 2y^2 -1$ . . . . Right!
6. PROBLEM 2:
$
2ab^4 - 3 a^2b^2 - ab^4 + a^2b^2 = ab^4 -2a^2b^2
$
You are correct.
7. PROBLEM 3:
$
-2(4y2+3z3+5) + 3(2y2-5z3+3) = (-8y^2 - 6z^3 - 10) + (6y^2 - 15z^3 + 9)$
$= -8y^2 - 6z^3 - 10 + 6y^2 - 15z^3 + 9 = -21z^3 - 2y^2 - 1$
Good job again. Problem 3 is correct.
-Andy
8. ## mystery equation
who had the mystery equation correct....
just curious..
9. Thanks for all of the help. I have a few more that I would like confirmation/correction for (exponents are denoted in red font). It helps to know if I am on the right track. 1) solve the equation x2 + 4x - 45 = 0 I have x=9 and x=-5 ??? 2) find the product of (x-2y)2 I have x2 - 4xy + 4x2 ??? 3) completely factor the expression y2 + 12y + 35 I have come up with (y+7) (y-5)?
10. BigWave, I think that it was Soroban. Thanks again!
11. Originally Posted by jay1
Thanks for all of the help. I have a few more that I would like confirmation/correction for (exponents are denoted in red font). It helps to know if I am on the right track. 1) solve the equation x2 + 4x - 45 = 0 I have x=9 and x=-5 ??? 2) find the product of (x-2y)2 I have x2 - 4xy + 4x2 ??? 3) completely factor the expression y2 + 12y + 35 I have come up with (y+7) (y-5)?
$x^2 + 4x - 45 = 0$ | {
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$x^2 + 4x - 45 = 0$
$(x+9)(x-5) = 0$
$x = 5, -9$
12. Originally Posted by jay1
2) Simplify the expression 2ab4 - 3 a2b2 - ab4 + a2b2 [COLOR=black]I have ab4 - 2a2b2
$
2ab^4 - 3a^2b^2 -ab^4 + a^2b^2 = ab^4 - 2a^2b^2
$
Atta boy (assuming you are a guy).
13. 3) completely factor the expression y2 + 12y + 35 I have come up with (y+7) (y-5)?
$
y^2 + 12y + 35 = (y+7)(y+5) \implies y=-5,-7
$
Good luck.
-Andy
14. Perform the indicated operations on this expression: (5a^3 + 3a -2) - (4a^3 + a^2 + 5) I have a^3 + a^2 + 3 Is this right?
15. Originally Posted by jay1
Perform the indicated operations on this expression: (5a^3 + 3a -2) - (4a^3 + a^2 + 5) I have a^3 + a^2 + 3 Is this right?
No.
$
(5a^3 + 3a -2) - (4a^3 + a^2 + 5) = 5a^3 + 3a - 2 - 4a^3 - a^2 - 5 =
a^3 - a^2 + 3a - 7
$
The minus before the parenthesis changes the sign of each term inside the parentheses.
-Andy
Page 1 of 2 12 Last | {
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How can I simplify this fraction problem?
I have the problem $$\frac{x^2}{x^2-4} - \frac{x+1}{x+2}$$ which should simplify to $$\frac{1}{x-2}$$
I have simplified $$x^2-4$$, which becomes:
$$\frac{x^2}{(x-2)(x+2)} - \frac{x+1}{x+2}$$
However, if I combine the fractions I get, $$x^2-x-1$$ for the numerator, which can't be factored. That's where I get stuck.
How can I get $$\frac{1}{x-2}$$ out of this problem?
You need to put them over a common denominator, $$\frac {x+1}{x+2}=\frac {(x+1)(x-2)}{(x+2)(x-2)}=\frac {x^2-x-2}{x^2-4}$$ Now you can subtract the numerators $$x^2-(x^2-x-2)=x+2$$ and finally divide out the $$x+2$$ from numerator and denominator
Your $$-1$$ should be $$-2$$. You didn't show your work, so I can't see why it happened.
• Regarding your last line, I wonder if OP just added the two numerators together without finding a common denominator. – Matthew Leingang Jan 30 at 18:54
• @MatthewLeingang That's right. It's been awhile since I worked with fractions, and I'm very rusty. – LuminousNutria Jan 30 at 18:56
• @LuminousNutria It's a very common error. But if you think about it with familiar fractions you'll remember it can't be true. For instance, $\frac{1}{2} + \frac{1}{2}$ is $1$, not $\frac{2}{4}$ (which is $\frac{1}{2}$ again). – Matthew Leingang Jan 30 at 19:00
\begin{align} \frac{x^2}{(x-2)(x+2)} - \frac{x+1}{x+2}&=\frac1{x+2}\left(\frac{x^2}{x-2} - (x+1)\right)\\ &=\frac1{x+2}\left(\frac{x^2-(x-2)(x+1)}{x-2}\right)\\ &=\frac1{x+2}\left(\frac{x^2-(x^2-x-2)}{x-2}\right)\\ &=\frac1{x+2}\left(\frac{x+2}{x-2}\right)\\ &=\frac1{x-2} \end{align}
Write $$\frac{x^2}{(x-2)(x+2)}-\frac{x+1}{x+2}=\frac{x^2}{(x-2)(x+2)}-\frac{(x+1)(x-2)}{(x+2)(x-2)}=…$$ Note that it must be $$x\ne 2,-2$$
The numerator should be $$x^2-(x-2)(x+1) = x+2$$ which simplifies with a factor of the denominator.
You probably just forgot a term when subtracting the two fractions. | {
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# Prove that for all integers $a$ and $b$, if $a$ divides $b$, then $a^{2}$ divides $b^{2}$.
I just need to know that if $a$ divides $b$, where $a$ and $b$ are integers, does $a^{2}$ divide $b^{2}$?
-
If $a$ divides $b$, then $b=ka$ for some integer $k$, so $b^2=k^2a^2$ where $k^2$ is an integer.
-
How do you prove that k^2 is an integer? Is that by closure? – amster27 Feb 22 '13 at 1:07
@amster27: If you are concerned about whether $k^{2}$ is an integer, then you should be equally concerned about whether both $a^{2}$ and $b^{2}$ are integers. :) – Haskell Curry Feb 22 '13 at 1:23
Thanks, I'm just starting to write proofs in a Mathematical Reasoning class, and I just feel so lost. But these explanations helped. – amster27 Feb 22 '13 at 1:28
@amster: Note that $+_{\mathbb{Z}}: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ and $\times_{\mathbb{Z}}: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$. Therefore, adding two integers or multiplying two integers yields an integer. – Haskell Curry Feb 22 '13 at 1:44
Hint $\rm\ \ a\mid b\ \Rightarrow\ \dfrac{b}a\in \Bbb Z\ \Rightarrow\ \dfrac{b^2}{a^2} = \left(\dfrac{b}a\right)^2\!\in \Bbb Z^2\subseteq \Bbb Z\:\Rightarrow\: a^2\mid b^2\$
-
That's closer to a proof than a hint. – 1015 Feb 22 '13 at 1:29
The argument is elegant, but I think that the OP wants to stay within $\mathbb{Z}$ and does not wish to jump into $\mathbb{Q}$, as he seems to be very interested in the axioms governing the properties of $\mathbb{Z}$. – Haskell Curry Feb 22 '13 at 1:29
@Haskell Possibly, but there was no hint of any such constraint in the question. – Math Gems Feb 22 '13 at 1:46 | {
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# Maximum number of squares touched by a line segment
Consider a square grid on the plane, with unit spacing. A line segment of integer length $$\L\$$ is dropped at an arbitrary position with arbitrary orientation. The segment is said to "touch" a square if it intersects the interior of the square (not just its border).
# The challenge
What is the maximum number of squares that the segment can touch, as a function of $$\L\$$?
# Examples
• L=3 $$\\ \, \$$ The answer is $$\7\$$, as illustrated by the blue segment in the left-hand side image (click for a larger view). The red and yellow segments only touch $$\6\$$ and $$\4\$$ squares respectively. The purple segment touches $$\0\$$ squares (only the interiors count).
• L=5 $$\\ \, \$$ The answer is $$\9\$$. The dark red segment in the right-hand side image touches $$\6\$$ squares (note that $$\5^2 = 3^2+4^2\$$), whereas the green one touches $$\8\$$. The light blue segment touches $$\9\$$ squares, which is the maximum for this $$\L\$$.
• The input $$\L\$$ is a positive integer.
• The algorithm should theoretically work for arbitrarily large $$\L\$$. In practice it is acceptable if the program is limited by time, memory, or data-type size.
• Input and output are flexible as usual. Programs or functions are allowed, in any programming language. Standard loopholes are forbidden.
• Shortest code in bytes wins.
# Test cases
Here are the outputs for L = 1, 2, ..., 50 (with L increasing left to right, then down):
3 5 7 8 9 11 12 14 15 17
18 19 21 22 24 25 27 28 29 31
32 34 35 36 38 39 41 42 43 45
46 48 49 51 52 53 55 56 58 59
60 62 63 65 66 68 69 70 72 73 | {
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• No OEIS entry, apparently (maybe soon) Jul 10 at 15:35
• I really liked this one from a problem-solving perspective. Jul 10 at 23:59
• My Math.SE question will be interesting to y'all and relevant to this question Jul 11 at 9:03
• This reminds me of the problem of optimizing beam placement in FTL Jul 13 at 2:02
• This sequence is now in OEIS: A346232 Jul 15 at 0:02
# Vyxal, 7 bytes
*d⇩√3+⌊
Try it Online!
* # Square
d # Double
⇩ # Subtract 2
√ # Square root
⌊ # Floor
# Python 2, 27 bytes
lambda L:(2*L*L-2)**.5//1+3
Try it online!
A direct formula:
$$f(L) = \lfloor \sqrt{2 L^2-2}\rfloor + 3$$
Derivation
As noted by @Kirill L. and others, the optimal layout uses a near-diagonal line segment whose horizontal and vertical span are at least $$\(h,h)\$$ or $$\(h,h+1)\$$. We need the length-$$\L\$$ to cover at least this much distance using the Pythagorean theorem, plus some extra to reach into squares and hit 3 more.
This gives a result of either:
• $$\2h+3\$$ where $$\h\$$ is the greatest positive integer where $$\h^2+h^2 < L^2\$$, or
• $$\2h+4\$$ where $$\h\$$ is the greatest positive integer where $$\h^2+(h+1)^2 ,
whichever of these is larger. We want to combine these two cases into one.
Let's start by making the two cases look more similar. Note that the second case can be rewritten as
$$\2(h+\frac{1}{2})+3\$$, where $$\2(h+\frac{1}{2})^2 + \frac{1}{2} < L^2\$$
or as
$$\2h+3\$$, where $$\2h^2 + \frac{1}{2} < L^2\$$
where $$\h\$$ is a positve integer-and-a-half. The $$\2h^2 in the first case can be also be written as $$\2h^2 +\frac{1}{2} < L^2\$$, which is equivalent because both sides were integers. So, the cases now merge into:
$$\2h+3\$$ where $$\h\$$ is the greatest positive integer or half-integer where $$\2h^2 +1/2 < L^2\$$
Calling $$\2h=H\$$, this is:
$$\H+3\$$ where $$\H\$$ is the greatest positive integer where $$\2(H/2)^2 +1/2 < L^2\$$. | {
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$$\H+3\$$ where $$\H\$$ is the greatest positive integer where $$\2(H/2)^2 +1/2 < L^2\$$.
This inequality is $$\H^2 < 2L^2-1\$$, and since these are integers, this is the same as $$\H^2 \leq 2L^2-2\$$. The greatest such positive integer $$\H\$$ is then $$\\lfloor \sqrt{2 L^2-2}\rfloor\$$, so the final result is $$\ \lfloor \sqrt{2 L^2-2}\rfloor + 3 \$$.
• Fantastic. "We want to combine these two cases into one. Let's start by making the two cases look more similar." Did you know somehow this could be done at the outset (if so, how?), or did you just guess it might be possible, and start playing around? Jul 11 at 3:55
• @Jonah It was a guess, but it seemed to me like something like this should be possible because the two types of outputs lie nearly on the same approximate curve.
– xnor
Jul 11 at 4:06
• Wonderful! Simpler than the formula I had obtained Jul 11 at 10:41
• I intend to submit this sequence to OEIS. Your formula is simpler than the one I had (i+j+3 with i = floor(L/sqrt(2)), j=ceil(sqrt(L^2-i^2))-1, so I'd like to give you credit (in addition to linking this challenge). How can I do that? If you prefer we can go on using e-mail Jul 11 at 15:22
• I may also have to write a short paper to support the submission, in particular to justify that the optimal segment orientation is always (h,h) or (h,h+1). Are interested in collaborating on this? Please let me know either way Jul 12 at 8:55
# R, 79 77 bytes
L=scan();j=1:L;a=j*2^.5;b=Mod(j+j*1i+1i);3+2*sum(L>a)+max(a[L>a],b[L>b])%in%b
Try it online!
Note: as it turned out, this is completely destroyed by xnor's formula, which would be 23 bytes in R:
(2*scan()^2-2)^.5%/%1+3
However, I'm keeping the existing code as a reference to my original solution.
### Original explanation | {
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