text stringlengths 2 132k | source dict |
|---|---|
It all depends on why you really want to compare the sets and what method makes a good comparison for what you want to do. Also, there are more ways of relating sets than what I have described here. Reply reply } Share Share New to Reddit? Create your account and connect with a world of communities. المواصلة باستخدام G... | {
"page_id": null,
"source": 6830,
"title": "from dpo"
} |
17 upvotes ·45 comments * * * * How is there an infinite amount of numbers between two numbers?. * * * 386K Members Online ### How is there an infinite amount of numbers between two numbers??]( yr. ago [r/AskEngineers]( Engineers apply the knowledge of math & science to design and manufacture maintainable systems used ... | {
"page_id": null,
"source": 6830,
"title": "from dpo"
} |
actual correct answer for 6÷2(1+2)?]( 142 upvotes ·158 comments * * * * Difference between 1E and 2E? Should I just wait for 2E if I'm just getting into the game?, is one more than the product of the numbers adjacent to that number (x-1)(x+1). Why is this, and are there any real-world applications using this principle?... | {
"page_id": null,
"source": 6830,
"title": "from dpo"
} |
r/maths]( yr. ago !Image 34: A banner for the subreddit^2]( 36: r/learnmath icon]( r/learnmath]( yr. ago !Image 37: r/learnmath icon. * * * 386K Members Online ### Difference between -5^2 and (-5)^2. * * * 386K Members Online [### What natural numbers really are?]( 5 upvotes ·14 comments * * * * [Why was 6 afraid of 7?... | {
"page_id": null,
"source": 6830,
"title": "from dpo"
} |
be justified in any possible way?]( 18 comments * * * * [What's 2+2?]( 45: r/patchouli_knowledge icon]( r/patchouli_knowledge]( mo. ago ![Image 46: A banner for the subreddit]( ![Image 47: r/patchouli_knowledge icon]( A subreddit dedicated to Patchouli Knowledge from the Touhou Project! * * * 1.1K Members Online [### W... | {
"page_id": null,
"source": 6830,
"title": "from dpo"
} |
* Are all integers fractions?. * * * 386K Members Online [### Are all integers fractions?]( 1 upvote ·7 comments * * * * [1 - 2 - 3 - 4… I DECLARE _______ WAR III]( 64: r/AskOuija icon]( r/AskOuija]( mo. ago ![Image 65: A banner for the subreddit]( ![Image 66: r/AskOuija icon]( r/AskReddit, Ouija-style. * * * 798K Memb... | {
"page_id": null,
"source": 6830,
"title": "from dpo"
} |
children at home (or considering it). Welcome to the largest home education subreddit! * * * 198K Members Online ### How to teach numbers 1 - 100 for 5 years old? * [Amazing]( * [Animals & Pets]( * [Cringe & Facepalm]( * [Funny]( * [Interesting]( * [Memes]( * [Oddly Satisfying]( * [Reddit Meta]( * [Wholesome & Heartwar... | {
"page_id": null,
"source": 6830,
"title": "from dpo"
} |
[Reality TV]( * [Romance Movies & Series]( * [Sci-Fi Movies & Series]( * [Superhero Movies & Series]( * [TV News & Discussion]( * * * * RESOURCES * [About Reddit]( * [Advertise]( * [Reddit Pro BETA]( * [Help]( * [Blog]( * [Careers]( * [Press]( * * * * [Communities]( * [Best of Reddit]( * [Topics]( ![Image 76]( | {
"page_id": null,
"source": 6830,
"title": "from dpo"
} |
Title: Surjective, injective and bijective linear maps URL Source: Markdown Content: In this lecture we define and study some common properties of linear maps, called surjectivity, injectivity and bijectivity. !Image 1: Infographic that helps to visualize the definitions of surjective, injective and bijective. coincid... | {
"page_id": null,
"source": 6834,
"title": "from dpo"
} |
themselves linear spaces (subspaces. Surjective map -------------- Let us start with a definition. Definition Let ![Image 22: $f$](blob: and ![Image 23: $f$](blob: be two linear spaces. Let ![Image 24: $f$](blob: be a linear map. The transformation ![Image 25: $f$](blob: is said to be surjective if and only if, for eve... | {
"page_id": null,
"source": 6834,
"title": "from dpo"
} |
by]( the vector![Image 54: $f$](blob: formally, we have that![Image 55: [eq10]]( are elements of ![Image 56: $f$](blob: that do not belong to ![Image 57: $f$](blob: For example, the vector ![Image 58: $f$](blob: not belong to ![Image 59: $f$](blob: because it is not a multiple of the vector ![Image 60: $f$](blob: the r... | {
"page_id": null,
"source": 6834,
"title": "from dpo"
} |
write the matrix product as a linear combination:![Image 85: $f$](blob: ![Image 86: $f$](blob: and ![Image 87: $f$](blob: are the two entries of ![Image 88: $f$](blob: Thus, the elements of ![Image 89: $f$](blob: are all the vectors that can be written as linear combinations of the first two vectors of the standard bas... | {
"page_id": null,
"source": 6834,
"title": "from dpo"
} |
spaces. A linear map ![Image 113: $f$](blob: is said to be bijective if and only if it is both surjective and injective. Solved exercises ---------------- Below you can find some exercises with explained solutions. ### Exercise 1 As we explained in the lecture on [linear maps]( a linear function ![Image 114: $f$](blob:... | {
"page_id": null,
"source": 6834,
"title": "from dpo"
} |
terms of a basis]( we have that!Image 145: $f$ !Image 150: $f$ it cannot be that both !Image 154: $f$. "Surjective, injective and bijective linear maps", Lectures on matrix algebra. | {
"page_id": null,
"source": 6834,
"title": "from dpo"
} |
Title: Can someone help me understand this? I’ve been thinking on it for days. : r/learnmath URL Source: Markdown Content: Can someone help me understand this? I’ve been thinking on it for days. : r/learnmath =============== Skip to main content. * * * 386K Members Online •5 yr. ago [owlbewatchinyou]( Can someone help... | {
"page_id": null,
"source": 6839,
"title": "from dpo"
} |
divided by 3. If I am wrong, or if someone has an explanation for this one, please let me know. My brain hurts! Read more Share Share [![Image 3: u/monday_com avatar]( The numbers don't lie - there's a reason why 10,000+ customers rate monday.com 5 stars and use it as their work management platform. It’s the #1 platfor... | {
"page_id": null,
"source": 6839,
"title": "from dpo"
} |
replies]( ![Image 7: u/RangerPL avatar _is_ equal to 10. Not "infinitely close" to 10 (whatever that means), but actually mathematically equal to 10. 10 can be divided by 3 - it's 3 1/3, which is a perfectly respectable number, with well-defined properties. If you are saying you can never fully write out 1/3 as a decim... | {
"page_id": null,
"source": 6839,
"title": "from dpo"
} |
explanation, I'm upvoting it because it is the one that helped the OP understand... Reply reply } Share Share 2 more replies 2 more replies More replies, right? Computers use it. In binary ten divided by three would be written as: # 1010 => one EIGHT plus no FOURS plus one TWO plus no UNITS 1010 / 11 but the result in ... | {
"page_id": null,
"source": 6839,
"title": "from dpo"
} |
in French or Chinese, etc. They all mean exactly the same number, just different ways of writing it down. Reply reply } Share Share 1 more reply 1 more reply More replies * * * 62K Members Online [### Anyone understands this?]( [![Image 14: r/AutismTranslated - Anyone understands this? ]( 65 upvotes ·84 comments * * * ... | {
"page_id": null,
"source": 6839,
"title": "from dpo"
} |
it here. Read the rules in the sticky. * * * 78K Members Online [### can someone calculate this for me]( [![Image 19: r/unexpectedfactorial - can someone calculate this for me]( 170 upvotes ·38 comments * * * * Promoted ![Image 20: sidebar promoted post thumbnail]( you guys verbose this?]( 21: r/IncreasinglyVerbose ico... | {
"page_id": null,
"source": 6839,
"title": "from dpo"
} |
work! Welcome to the fastest growing color grading community! * * * 26K Members Online ### Can anyone explain this!, and our rules are designed to reinforce this. * * * 668K Members Online [### [5th grade math]. I would think 300 is 100 times less than 30,000. Am I thinking about this wrong?]( [![Image 37: r/HomeworkHe... | {
"page_id": null,
"source": 6839,
"title": "from dpo"
} |
to problem-solving in mathematics across various topics and levels. * * * 59K Members Online ### Struggling with this question. Would appreciate you guys help :)]( 34 upvotes ·62 comments * * * * What do you call the numbers to the right of the decimal point? and its use around the world. Posts discussing metrology, me... | {
"page_id": null,
"source": 6839,
"title": "from dpo"
} |
for questions of a mathematical nature. Please read the subreddit rules below before posting. * * * 198K Members Online ### I’m confuse on how did they get 2/3 + 2/square 13 to 26+6 square root 13/39? I get everything else except that part.. * * * 386K Members Online ### Best way to multiply large numbers?. * * * 386K ... | {
"page_id": null,
"source": 6839,
"title": "from dpo"
} |
comments * * * * [[elementary school math? lol] What does it mean to divide something?]( 66: r/learnmath icon]( r/learnmath]( yr. ago !Image 67: r/learnmath icon. * * * 386K Members Online [### [elementary school math? lol] What does it mean to divide something?]( 11 upvotes ·9 comments * * * * [Why is 0/0 undefined an... | {
"page_id": null,
"source": 6839,
"title": "from dpo"
} |
here if you happen to live in a universe which follows the 'laws of physics'.* * * * 60K Members Online ### Can anyone see why this doesn’t work * * * 77K Members Online ### This makes a lot of sense to me. * [Amazing]( * [Animals & Pets]( * [Cringe & Facepalm]( * [Funny]( * [Interesting]( * [Memes]( * [Oddly Satisfyin... | {
"page_id": null,
"source": 6839,
"title": "from dpo"
} |
* [Strategy Games]( * [Tabletop Games]( * Q&As * [Q&As]( * [Stories & Confessions]( * Technology * [3D Printing]( * [Artificial Intelligence & Machine Learning]( * [Computers & Hardware]( * [Consumer Electronics]( * [DIY Electronics]( * [Programming]( * [Software & Apps]( * [Streaming Services]( * [Tech News & Discussi... | {
"page_id": null,
"source": 6839,
"title": "from dpo"
} |
Title: Just a moment... URL Source: Warning: Target URL returned error 403: Forbidden Warning: This page maybe requiring CAPTCHA, please make sure you are authorized to access this page. Markdown Content: academia.stackexchange.com -------------------------- Verify you are human by completing the action below. academi... | {
"page_id": null,
"source": 6841,
"title": "from dpo"
} |
Title: Just a moment... URL Source: Warning: Target URL returned error 403: Forbidden Warning: This page maybe requiring CAPTCHA, please make sure you are authorized to access this page. Markdown Content: mathoverflow.net ---------------- Verify you are human by completing the action below. mathoverflow.net needs to r... | {
"page_id": null,
"source": 6841,
"title": "from dpo"
} |
Title: unsloth/OpenMathReasoning · Datasets at Hugging Face URL Source: Markdown Content: Subset (1) default·193k rows Split (1) | expected_answer string lengths 1 794 | problem_type string classes 1 value | problem_source string classes 5 values | generation_model string classes 1 value | pass_rate_72b_tir string cla... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
x \). Let's tackle each case one by one. **Analyzing Case 1: \( x \geq 0 \)** We need to find the number of real roots of \( f(x) = x^6 + x - 7 \) where \( x \geq 0 \). Let me think about the behavior of this function. For \( x = 0 \): \[ f(0) = 0 + 0 - 7 = -7 \] So at x=0, f(x) is -7. As \( x \) increases, the \( x^6 ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
\( x \) approaches negative infinity and see if there's a crossing. First, when \( x \) is negative and approaches negative infinity, \( x^6 \) is still positive and grows rapidly (since even power), while the term -x becomes positive because x is negative. So \( x^6 \) and \( -x \) are both positive, so as \( x \to -\... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
real roots: one positive and one negative. But wait, is that correct? Let me double-check. In Case 1, yes, there's one positive root. In Case 2, one negative root. So total of two real roots. Therefore, the answer should be 2. But wait, hold on. Let me try plugging in some numbers to verify. For the positive case, let'... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
decreases from 59 to -5, crossing zero somewhere in between. So that's one real root between x=-2 and x=-1. Wait, but what about from x=-infty to x=-2? If the function is approaching infinity as x approaches -infty, and at x=-2, it's 59, which is still positive. So the function is decreasing from infinity to 59 as x go... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
it's still positive. Then between x=-1.5 and x=-1, it goes from ~5.89 to -5, crossing zero once. So only one root in the negative case. Therefore, total two real solutions. So the answer is 2. So the number of real numbers satisfying the equation is 2. But just to be thorough, let me check if there could be any other r... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
\) at the endpoints: \[ f(0) = 0^6 + 0 - 7 = -7 \] As \( x \to \infty \), \( x^6 \) grows much faster than \( x \), so \( f(x) \to \infty \). By the Intermediate Value Theorem, since \( f(x) \) is continuous and strictly increasing from \(-7\) to \(\infty\), there is exactly one root in the interval \( [0, \infty) \). ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
= 7 \) is \(\boxed{2}\). | cot | | 11 | has_answer_extracted | aops_c4_high_school_math | DeepSeek-R1 | 0.96875 | For how many positive integer values of $n$ is $n^3+1631$ divisible by $n+11$? | Okay, so I need to find how many positive integer values of n make n³ + 1631 divisible by n + 11. Hmm, divisibility here mean... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
no solutions? But the problem is asking for positive integer values of n. That seems contradictory. Maybe I'm applying the theorem incorrectly? Wait, no. The Remainder Theorem tells us that the remainder when dividing by n + 11 is f(-11). So, if n + 11 divides f(n), then the remainder must be zero, which would require ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
is 2² × 3 × 5². The divisors can be generated by exponents of these primes: 2^0, 2^1, 2^2; 3^0, 3^1; 5^0, 5^1, 5^2. So the total number of divisors is (2+1)(1+1)(2+1) = 3*2*3 = 18 divisors. Let me list all divisors: Starting from 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60, 75, 100, 150, 300. Wait, maybe not in or... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
11 = 19 d = 50: 50 - 11 = 39 d = 60: 60 - 11 = 49 d = 75: 75 - 11 = 64 d = 100: 100 - 11 = 89 d = 150: 150 - 11 = 139 d = 300: 300 - 11 = 289 So these would be the possible n's: 1, 4, 9, 14, 19, 39, 49, 64, 89, 139, 289. Let's count them: 11 values. Therefore, there are 11 positive integer values of n. But wait, let me... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
even though calculating 289³ is tedious, we can trust the logic here. Therefore, the answer should be 11 positive integer values of n. So the number of such n is 11. Therefore, the final answer is 11. Wait, but just to confirm, let's take another example where d is smaller. Suppose n +11=12, which gives n=1. Then 1^3 +... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
to 12 are: \[ 12, 15, 20, 25, 30, 50, 60, 75, 100, 150, 300 \] Each of these divisors \( d \) corresponds to a positive integer \( n = d - 11 \). Calculating these values: \[ \begin{align*} 12 - 11 &= 1, \\ 15 - 11 &= 4, \\ 20 - 11 &= 9, \\ 25 - 11 &= 14, \\ 30 - 11 &= 19, \\ 50 - 11 &= 39, \\ 60 - 11 &= 49, \\ 75 - 11... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
with the equation given. Let me rewrite the equation: b*(10a + c) = c*(10a + b) + 50 Expanding both sides: 10ab + bc = 10ac + bc + 50 Wait, let me check that. On the left side, when we expand b*(10a + c), we get 10ab + bc. On the right side, c*(10a + b) is 10ac + bc. Then adding 50 gives 10ac + bc + 50. So subtract the... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
= -5, (b - c) = -1: invalid. Alternatively, maybe considering fractions? But since a is a digit (integer from 1-9), and (b - c) must also be an integer (since b and c are digits), so only integer solutions. Thus, only two possibilities: 1) a = 1, b - c = 5 2) a = 5, b - c = 1 But wait, let's check if these are the only... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
5. So c can be from 0 to 4, and b from 5 to 9. Case 2: a = 5, then (b - c) = 1. So b = c + 1. So c can be from 0 to 8, and b from 1 to 9. Any other cases? For example, a = 5, (b - c) = 1. If a were 5, and (b - c) =1. Or a=1, (b - c)=5. Since 5 is prime, there are no other integer solutions where a is a digit (1-9) and ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
let's check the original equation with a=5, b=9, c=8. Left side: b*(10a + c) = 9*(50 +8) = 9*58 = 522 Right side: c*(10a + b) +50 =8*(50 +9) +50 =8*59 +50=472 +50=522. So yes, 522=522. So 598 is a valid number. But wait, the number is 598, but in this case, a=5, b=9, c=8. So 598 is a valid number. Is there a larger num... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
as we have. If a were 5 and (b -c)=1, then possible numbers are from 510 (c=0, b=1) up to 598 (c=8, b=9). So 598 is the maximum in that case. Alternatively, is there a way to get a higher a? For example, a=6. Then, if a=6, (b - c)=5/6, which is not an integer. Similarly, a=2, (b - c)=5/2=2.5, which is not an integer. S... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
\(b - c\) such that their product is 5 are: 1. \(a = 1\) and \(b - c = 5\) 2. \(a = 5\) and \(b - c = 1\) **Case 1: \(a = 1\) and \(b - c = 5\)** Here, \(b = c + 5\). The maximum value for \(c\) is 4 (since \(b\) must be a digit), leading to \(b = 9\). The number is 194. Checking the equation: - Left side: \(9 \cdot (1... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
divisible by 12. But I'm a bit fuzzy on the exact steps. Let me start by recalling some basic concepts. First, the general form of a linear congruence is ax ≡ b mod m. In this case, a is 3, b is 5, and m is 12. To solve for x, I think the first thing to check is whether 3 and 12 are coprime, which means their greatest ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
there's no solution. Therefore, the congruence equation 3x ≡ 5 mod 12 has no solution. Wait, but let me check with some numbers. If I plug in some x values and see if 3x mod 12 equals 5. Let's try x=1: 3*1=3 mod12 → 3. Not 5. x=2: 6 mod12=6. x=3:9. x=4:12 mod12=0. x=5:15 mod12=3. x=6:18 mod12=6. x=7:21 mod12=9. x=8:24 ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
represented? Sometimes in congruence problems, if there's no solution, you might state that there is no solution. But maybe in the context of modular equations, you can write "No solution exists" inside the box. Alternatively, if they expect the answer in a particular format. But the problem says "Solve the congruence ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
mod 4, there's no solution because gcd(2,4)=2, which doesn't divide 1. So the answer would be no solution. Therefore, similarly here, it's no solution. Therefore, the final answer is \boxed{\text{No solution}}. To solve the congruence equation \(3x \equiv 5 \pmod{12}\), we need to determine if there exists an integer \... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
| DeepSeek-R1 | 0.96875 | How many positive divisors of \(2^7 \cdot 3^5 \cdot 5^3 \cdot 7^2\) are divisible by 10? | Okay, so I need to find how many positive divisors of the number \(2^7 \cdot 3^5 \cdot 5^3 \cdot 7^2\) are divisible by 10. Hmm, let's start by recalling that when dealing with divisors of a number in it... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
(so 3 choices instead of 4), and for 3 and 7, they can still be from 0 up to their maximums (since they don't affect divisibility by 10). So the number of such divisors should be the product of the number of choices for each exponent. For 2: 7 choices (1-7), for 5: 3 choices (1-3), for 3: 6 choices (0-5), and for 7: 3 ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
5^3 \cdot 7^2\) is (7+1)(5+1)(3+1)(2+1) = 8*6*4*3 = let's compute that: 8*6=48, 48*4=192, 192*3=576. So total divisors are 576. But we want those divisible by 10, which is 2*5, so we need to fix the exponents of 2 and 5 to be at least 1. So, when you fix an exponent to be at least 1, you subtract 1 from the original ex... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
2 and 5. Alternatively, since 10 is 2*5, the number of divisors divisible by 10 is the same as the number of divisors divisible by both 2 and 5. Which is equal to the number of divisors that have at least one 2 and at least one 5. Which is exactly the first method's answer. Alternatively, using inclusion-exclusion: tot... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
both methods confirm the same answer. Therefore, I think 378 is correct. Let me verify once more. Original number N has prime factors 2^7, 3^5, 5^3, 7^2. To find divisors divisible by 10, which is 2*5. So such a divisor must have at least 2^1 and 5^1. So when constructing such a divisor, the exponents for 2 can be 1-7 ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
378 \] Thus, the number of positive divisors of \(2^7 \cdot 3^5 \cdot 5^3 \cdot 7^2\) that are divisible by 10 is \(\boxed{378}\). | cot | | \(\frac{1}{2}x^2 - \frac{3}{2} \ln(x^2 + 3) + \frac{1}{\sqrt{3}} \arctan \left( \frac{x}{\sqrt{3}} \right) + C\) | has_answer_extracted | aops_c7_college_math | DeepSeek-R1 | 0.96... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
C. But let's not add the constant yet since we're dealing with an indefinite integral and will combine constants at the end. Now, the second integral is ∫(-3x + 1)/(x² + 3) dx. This can be split into two parts: -3∫x/(x² + 3) dx + ∫1/(x² + 3) dx. Let's handle each part separately. Starting with the first part: -3∫x/(x² ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
dx\), we start by performing polynomial long division on the rational function \(\frac{x^3 + 1}{x^2 + 3}\). 1. **Polynomial Long Division:** - Divide \(x^3 + 1\) by \(x^2 + 3\). - The quotient is \(x\) and the remainder is \(-3x + 1\). - Thus, we can write: \[ \frac{x^3 + 1}{x^2 + 3} = x + \frac{-3x + 1}{x^2 + 3}. \] 2... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
cot | | 3 | has_answer_extracted | aops_c6_high_school_olympiads | DeepSeek-R1 | 0.96875 | Find the largest positive integer \( n \) such that the equations \((x + 1)^2 + y_1^2 = (x + 2)^2 + y_2^2 = \cdots = (x + n)^2 + y_n^2\) have integer solutions in \( x, y_1, y_2, \ldots, y_n \). | Okay, let's tackle this problem.... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
Pythagorean triple. The challenge is to find the maximum \( n \) for which such an \( x \) and \( k \) exist. Let me consider the differences between consecutive terms. For example, the difference between \( (x + i + 1)^2 + y_{i+1}^2 \) and \( (x + i)^2 + y_i^2 \) must be zero since they are all equal. So: \[ (x + i + ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
\), which implies that \( k \) must be greater than \( x + i \), so \( k \) must be greater than \( x + n \). Alternatively, thinking of each \( x + i \) as one leg of a right triangle with hypotenuse \( k \). So, each \( x + i \) must be less than \( k \), and the other leg \( y_i \) is then determined. But how can mu... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
with the same hypotenuse \( k \). I need to recall or research if there is any known result about such hypotenuses. Alternatively, maybe construct an example. Alternatively, perhaps parametrize the problem. Suppose that for each \( i \), \( (x + i, y_i, k) \) is a Pythagorean triple. So: \( (x + i)^2 + y_i^2 = k^2 \) W... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
\( \gcd(a_i, b_i) = d \), then \( a_i = d m_i^2 \), \( b_i = d n_i^2 \), where \( m_i \) and \( n_i \) are coprime integers. Then, \( d(m_i^2 + n_i^2) = 2k \), and \( d(n_i^2 - m_i^2) = 2x + 2i \). So for each \( i \), these equations must hold. Therefore, \( d \) must divide both \( 2k \) and \( 2x + 2i \). Since this... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
\( x + i = 2 m_i n_i \), and \( y_i = m_i^2 - n_i^2 \). However, since all the \( k \) must be the same, but the parameters \( m_i \) and \( n_i \) can vary for each \( i \). But if we need all these triples to share the same hypotenuse \( k \), then \( k \) must be expressible as \( m_i^2 + n_i^2 \) in multiple ways. ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
y_1^2) = 0 \) So: \( y_2^2 - y_1^2 = -2x - 3 \) Factor left side: \( (y_2 - y_1)(y_2 + y_1) = -2x - 3 \) Since \( x \) is an integer, the RHS must be an integer. Let's let \( d_1 = y_2 - y_1 \), \( d_2 = y_2 + y_1 \), so \( d_1 d_2 = -2x -3 \). Also, \( d_1 \) and \( d_2 \) have the same parity since \( y_2 \) and \( y... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
+ d_2 \) must be even, but since both \( d_1 \) and \( d_2 \) are odd, their sum is even, so that's okay. So, suppose we pick some factors. For example, let's take \( -2x -3 = (-1)(2x + 3) \), so factor pairs like (-1, 2x + 3), (-3, (2x + 3)/3), etc. But perhaps we can find a concrete example. Let me pick \( d_1 = -1 \... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
\): Then, first equation: \( 1^2 + 2^2 = 1 + 4 = 5 = k^2 \), which is not a perfect square. So, \( k = \sqrt{5} \), not integer. So, this approach doesn't work. Alternatively, perhaps pick another factorization. Let's take \( -2x -3 = (-3)(something) \). Let's suppose \( d_1 = -3 \), \( d_2 = ( -2x -3 ) / (-3) = (2x + ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
\). If \( t = 0 \), then \( k^2 = 5 \), not a square. \( t = 1 \): 10 + 10 + 5 = 25, which is 5^2. So, \( k = 5 \). Then, x = 3*1 = 3. Check the equations: First equation: (3 + 1)^2 + y_1^2 = 16 + y_1^2 = 25 ⇒ y_1^2 = 9 ⇒ y_1 = 3 or -3. But from above, y_1 = t + 2 = 1 + 2 = 3. So, yes. Second equation: (3 + 2)^2 + y_2^... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
y2^2 - y1^2 = 0 Which is 2x + 3 + (y2 - y1)(y2 + y1) = 0 Second subtraction: (x + 3)^2 - (x + 2)^2 + y3^2 - y2^2 = 0 Which is 2x + 5 + (y3 - y2)(y3 + y2) = 0 So, similar to before, but now we have two equations: 1. (y2 - y1)(y2 + y1) = -2x -3 2. (y3 - y2)(y3 + y2) = -2x -5 Hmm, so similar structure. Let's see if we can... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
65^2? Let's compute: 16^2=256, 63^2=3969, sum 4225. 65^2=4225, yes. Also, 33^2 + 56^2 = 1089 + 3136 = 4225, so 33 and 56. Also, 35^2 + 60^2 = 1225 + 3600 = 4825, which is not 65^2. Wait, 65^2 is 4225, so 63 and 16, 33 and 56, etc. Are there three consecutive numbers in 16, 33, 56, 63? No. Alternatively, maybe k=85. Pyt... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
+4^2= 31^2 +16=961 +16=977. Wait, no. Wait, 31^2 +4^2=961 +16=977, which is not 1105. Wait, that's not correct. Wait, 1105 = 33^2 +4^2=1089 +16=1105. 32^2 +9^2=1024 +81=1105. 31^2 +12^2=961 +144=1105. So yes, they are correct. So if we set x=30, then: x +1=31, y1=4: 31^2 +4^2=1105=k^2 x +2=32, y2=9:32^2 +9^2=1105=k^2 x... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
just an integer, which can be written as a sum of two squares for each i. Therefore, it's possible that the common value is a number that can be written as a sum of two squares in n different ways, with the first squares being consecutive integers x+1, x+2,...,x+n. But in that case, maybe we can have larger n. But in t... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
known to be 1^2 + 18^2, 6^2 + 17^2, 10^2 + 15^2, 15^2 + 10^2, etc. But the consecutive numbers here are 1,6,10,15... not consecutive. Another example, 625 is 0^2 +25^2, 7^2 +24^2, 15^2 +20^2. Again, not consecutive. What about 50? 1^2 +7^2=50, 5^2 +5^2=50, 7^2 +1^2=50. No. Wait, perhaps a larger number. Let's consider ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
+1)^2 = - (2x + 2i +1) So, similar to earlier, we have: y_{i+1}^2 - y_i^2 = - (2x + 2i +1 ) Which is the same equation as before. Therefore, this leads us back to the same recurrence relation. So, this is a system of n-1 equations: For each 1 ≤ i ≤ n-1: y_{i+1}^2 - y_i^2 = - (2x + 2i +1 ) Which can be rewritten as: (y_... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
each equation. So, the challenge is to find an x and a set of d_i, e_i that satisfy all these equations. Let me consider small n first. For n=3, we had: For i=1: d_1 * e_1 = - (2x + 3 ) For i=2: d_2 * e_2 = - (2x + 5 ) If we consider the previous example where x=30 and S=1105, then: For i=1: d_1 * e_1 = - (2*30 + 3 )= ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
let me correct that. For hypotenuse 1105: 31^2 + 12^2 = 961 + 144 = 1105 32^2 + 9^2 = 1024 + 81 = 1105 33^2 +4^2=1089 +16=1105 So x +1=31, x +2=32, x +3=33 ⇒ x=30. Then y1=12, y2=9, y3=4. Ah, okay, I had the y_i's backwards before. So, the previous equations were: For i=1: y2^2 - y1^2 = 9^2 -12^2 =81 -144= -63= - (2x +... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
want four consecutive integers a, a+1, a+2, a+3 such that each can be part of a Pythagorean triple with the same hypotenuse S. So: a^2 + b^2 = S (a+1)^2 + c^2 = S (a+2)^2 + d^2 = S (a+3)^2 + e^2 = S Subtracting consecutive equations: For a and a+1: (a+1)^2 - a^2 + c^2 - b^2 =0 Which is 2a +1 + (c - b)(c + b)=0 Similarl... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
Let me try to express all variables in terms of x. For the first equation, suppose we take y2 - y1 = -1, y2 + y1 = 2x +3 Then, solving for y1 and y2: Adding equations: 2y2 = (2x +3) -1 ⇒ 2y2 = 2x +2 ⇒ y2 =x +1 Subtracting: 2y1 = (2x +3) +1 ⇒ 2y1 =2x +4 ⇒ y1 =x +2 So, this gives y1 =x +2, y2=x +1. Substituting into the ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
if d=1, e=2x +5: Then y3 - y2= -1, y3 + y2=2x +5. Then, adding: 2y3=2x +4 ⇒ y3=x +2 Subtracting: 2y2=2x +6 ⇒ y2=x +3 But earlier, from the first equation, y2=x +1. So x +3= x +1 ⇒ impossible. Alternatively, d=5, e=(2x +5)/5. Then: y3 - y2= -5, y3 + y2= (2x +5)/5. But then adding: 2y3= (2x +5)/5 -5 ⇒ y3= (2x +5 -25)/10=... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
y1 t +9 t^2) - (y1^2 +4 y1 t +4 t^2)=2 y1 t +5 t^2= - (2x +7 ) Now, set up the system: 1. 2 y1 t + t^2 = - (2x +3 ) 2. 2 y1 t +3 t^2 = - (2x +5 ) 3. 2 y1 t +5 t^2 = - (2x +7 ) Subtract the first equation from the second: (2 y1 t +3 t^2) - (2 y1 t +t^2)= 2 t^2 = - (2x +5 ) + (2x +3 )= -2 Therefore: 2 t^2 = -2 ⇒ t^2 = -1... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
is getting quite involved. Perhaps trying specific small values for y1, y2, etc., to see if we can find integers that satisfy these equations. Alternatively, think of this as a system of Diophantine equations. Suppose we set y4 = y3 - a, and y3 = y2 - b, y2 = y1 - c. Then substituting backwards, but this might not lead... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
is even or odd. Let's suppose a is even. Let me first assume that (c -b) = -1, then (c +b) = 2a +1. Then: Adding: 2c = (2a +1) -1 ⇒2c=2a ⇒c=a Subtracting: 2b = (2a +1) +1 ⇒2b=2a +2⇒b=a +1 So, c=a, b=a +1. Then, S= a^2 + (a +1)^2=2a^2 +2a +1. Then, from the second equation: S=(a+1)^2 +c^2=(a+1)^2 +a^2=2a^2 +2a +1, which... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
+d= -4 Adding: 2(a -1)= -5 ⇒a -1= -2.5 ⇒a= -1.5, invalid. Case 4: a -1 -d= -2 a -1 +d= -2 Adding: 2(a -1)= -4 ⇒a -1= -2 ⇒a= -1, invalid. So, no solution in this case. Therefore, assuming (c -b)= -1 leads to no valid solution. Let's try another factor pair for the first equation. Suppose (c -b)= -k, (c +b)= (2a +1)/k. F... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
m= -1, n=6t +5. Then: d - (t -1)= -1 ⇒d= t -2 d +t -1=6t +5 ⇒d=6t +5 -t +1=5t +6 But d= t -2=5t +6 ⇒t -2=5t +6 ⇒-4t=8 ⇒t= -2. Then, a=3*(-2)+1= -5, invalid. Alternatively, take m=1, n= - (6t +5 ). Then: d - (t -1)=1 ⇒d= t d +t -1= - (6t +5 ) ⇒t +t -1= -6t -5 ⇒2t -1= -6t -5 ⇒8t= -4 ⇒t= -0.5, invalid. Alternatively, take... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
(5+1)^2 +c^2=S ⇒36 +c^2=25 +b^2⇒b^2 -c^2=11. Factors of11: (1,11) or (-1,-11). So: b -c=1, b +c=11 ⇒b=6, c=5. Then S=25 +36=61. Next, (5+2)^2 +d^2=49 +d^2=61 ⇒d^2=12, not integer. So no. Alternatively, a=2. S=4 +b^2. Next equation:9 +c^2=4 +b^2⇒b^2 -c^2=5. Factors:1,5 or -1,-5. b -c=1, b +c=5 ⇒b=3, c=2. S=4+9=13. Next ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
as: 1470= 1470^2 + 0^2 But not helpful. Alternatively, S=325=1^2 + 18^2=6^2 + 17^2=10^2 + 15^2=15^2 +10^2=17^2 +6^2=18^2 +1^2. So legs 1,6,10,15,17,18. No consecutive numbers. Alternatively, S=221=10^2 +11^2=5^2 +14^2. 10 and 11 are consecutive, so x=9, y1=11, y2=14. But check: (10)^2 +11^2=221, (11)^2 +14^2=121+196=31... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
= 0 \] \[ 2x + 2i + 1 + y_{i+1}^2 - y_i^2 = 0 \] \[ y_{i+1}^2 - y_i^2 = - (2x + 2i + 1) \] This can be factored as: \[ (y_{i+1} - y_i)(y_{i+1} + y_i) = - (2x + 2i + 1) \] 2. **Example for \( n = 3 \)**: - Consider the number \( S = 1105 \): \[ 31^2 + 12^2 = 1105 \] \[ 32^2 + 9^2 = 1105 \] \[ 33^2 + 4^2 = 1105 \] - Here... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
\). Thus, the largest positive integer \( n \) is \(\boxed{3}\). | cot | | \( P(x) = cx \) | has_answer_extracted | aops_c4_high_school_math | DeepSeek-R1 | 0.96875 | Find all polynomials \( P \in \mathbb{R}[x] \) such that: 1) \( P(0) = 0 \) 2) \( 2P(x) = P(x - 1) + P(x + 1) \) for all \( x \in \mathbb{R} \) | Okay, l... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
- 2P(x) + P(x - 1) = 0. This resembles the second difference operator. In finite differences, the second difference of a function is defined as Δ²P(x) = Δ(ΔP(x)) where ΔP(x) = P(x + 1) - P(x). So, the second difference Δ²P(x) would be [P(x + 2) - P(x + 1)] - [P(x + 1) - P(x)] = P(x + 2) - 2P(x + 1) + P(x). But our equa... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
we have the additional condition that P(0) = 0. So for a linear polynomial, P(0) = 0 implies that the constant term is zero, which is already satisfied by P(x) = ax. So linear polynomials of the form ax satisfy both conditions. But the problem says "find all polynomials," so maybe linear polynomials are the only soluti... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
Compute P(x + 1) = a(x + 1)³ + b(x + 1)² + c(x + 1) = a(x³ + 3x² + 3x + 1) + b(x² + 2x + 1) + c(x + 1) = ax³ + 3ax² + 3ax + a + bx² + 2bx + b + cx + c. Similarly, P(x - 1) = a(x - 1)³ + b(x - 1)² + c(x - 1) = a(x³ - 3x² + 3x - 1) + b(x² - 2x + 1) + c(x - 1) = ax³ - 3ax² + 3ax - a + bx² - 2bx + b + cx - c. Adding P(x + ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
0 2b = 0 ⇒ b = 0 So, if a = 0 and b = 0, then the polynomial becomes P(x) = cx, which is linear. So again, the cubic case reduces to the linear solution. Therefore, cubics don't add any new solutions. Similarly, maybe for any higher degree polynomial, when we compute P(x + 1) + P(x - 1) - 2P(x), we end up with equation... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
+ b - (ax + b) = a. So if Q(x) is constant, then P(x) must be linear. However, since P(0) = 0, then b = 0, so P(x) = ax. Therefore, the only solutions are linear polynomials passing through the origin. But let me check that conclusion again. If Q(x) is a constant, then integrating (summing) the differences, we get that... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
Since the problem asks to put the answer inside \boxed{}, probably boxed as \boxed{P(x) = kx} where k is a real constant. However, in some contexts, they might prefer to write it as \boxed{0} if k must be zero, but in this case, our analysis shows that any linear polynomial with P(0) = 0 works. Let me just confirm once... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
included when a=0. So yes, all P(x) = a x. So the final answer is \boxed{a x} with a ∈ ℝ. But the problem says to "put the answer (and only answer) inside \boxed{}", so perhaps stating the general form. Since in many math problems, they write the answer as \boxed{P(x) = ax} where a is a real constant. But if they want ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
Consider the functional equation \( 2P(x) = P(x - 1) + P(x + 1) \). We can rewrite this as: \[ P(x + 1) - 2P(x) + P(x - 1) = 0. \] - This equation resembles the second difference of the polynomial \( P(x) \). For a polynomial \( P(x) \) of degree \( n \), the second difference \( \Delta^2 P(x) \) is a polynomial of deg... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
1) = (ax^2 + 2ax + a + bx + b) + (ax^2 - 2ax + a + bx - b) = 2ax^2 + 2bx + 2a. \] According to the functional equation: \[ 2P(x) = 2(ax^2 + bx) = 2ax^2 + 2bx. \] Equating the two expressions: \[ 2ax^2 + 2bx + 2a = 2ax^2 + 2bx. \] This implies \( 2a = 0 \), so \( a = 0 \). Thus, \( P(x) \) reduces to a linear polynomial... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
1 x² = 1 + 2001 = 2002 2002 is not a perfect square. The square of 44 is 1936 and 45² is 2025, so 2002 is in between, so no solution here. y=2: y! = 2 x² = 2 + 2001 = 2003 Again, 44²=1936, 45²=2025, so 2003 isn't a square. No solution. y=3: y! = 6 x² = 6 + 2001 = 2007 Same range, between 44² and 45². Not a square. y=4:... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
So no square. So, as y increases, y! becomes way too big, and x² would need to be even larger, but even then, since the gap between consecutive squares increases, the chance that y! +2001 is a perfect square becomes very low. But perhaps we should check up to y=7 or 8. But maybe there's another solution. Wait, let's ch... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.