id string | solution string | answer string | metadata dict | problem string | tags list | idea string | original_index int64 | candidates list | bma_scores list | jaccard_scores list | domains list | relevance_scores list | golden_tags list |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
aops_1270027 | [quote=dogofmath]No because it is $2 \mod{5}$ so it cannot be a perfect square.[/quote]
If anybody is wondering why any number that is $2\pmod{5}$ can't be a perfect square, here's why. A number that is a multiple of $5$ is either ends in a $5$ or $0$. So a number $2\pmod{5}$ either ends with a $7$ or $2$, but all per... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Is $2001^{326}+326^{2001}$ a perfect square?",
"content_html": "Is <img src=\"//latex.artofproblemsolving.com/7/b/f/7bf423920433ab101454e0b841dba97728af898a.png\" class=\"latex\" alt=\"$2001^{326}+32... | Is \(2001^{326} + 326^{2001}\) a perfect square? | [
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"/Mathematics/NumberTheory/Congruences/Modulus",
"/Mathematics/NumberTheory/Congruences/Resi... | Compute the sum modulo 5, noting squares are only 0,1,4; since it’s 2 mod 5, it cannot be a perfect square. | 152,726 | [
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"/Mathematics/Algebra/AlgebraicOperations/GeneralAlgebraicOperations/NaturalHomomorphism",
"/Mathematics/Algebra/AlgebraicOperations/Gen... |
aops_395888 | [quote="AndrewTom"]Given that $I_{n} = \int_{0}^{1} (1+x^{2})^{n} dx$, show that
$(2n+1)I_{n}= 2^{n} + 2nI_{n-1}$.
[/quote]
[hide="show recurrence"]
Using parts, let $u=(1+x^{2})^{n}, \;\ dv=dx, \;\ du=2nx(1+x^{2})^{n-1}dx, \;\ dv=x$
$x(1+x^{2})^{n}|_{0}^{1} -2n\int_{0}^{1}x^{2}(1+x^{2})^{n-1}dx$
The left side ev... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Given that $I_{n} = \\int_{0}^{1} (1+x^{2})^{n} dx$, show that\n\n$(2n+1)I_{n}= 2^{n} + 2nI_{n-1}$.\n\n(a) Verify that the method used to obtain the reducion formula is valid for all real values of $n$.\n\... | Given \(I_{n}=\displaystyle\int_{0}^{1}(1+x^{2})^{n}\,dx\), show that
\[
(2n+1)I_{n}=2^{n}+2nI_{n-1}.
\]
(a) Verify that the method used to obtain the reduction formula is valid for all real values of \(n\).
(b) What form does the reduction formula \((2n+1)I_{n}=2^{n}+2nI_{n-1}\) take when \(n=0\)? Check that this is... | [
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"/Mathematics/CalculusandAnalysis/Calculus/Integrals/DefiniteIntegrals"
] | Integrate by parts then rewrite the x² factor as (1+x²)‑1 to express the remaining integral in terms of Iₙ and Iₙ₋₁. | 248,230 | [
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aops_1628862 | [quote=Yimself]the standard substitution $tan(x/2)=u$ after splitting the integral from 0 to pi and from pi to 2pi (because tangent is not continous there) Will this work for you?[/quote]
Yes but first write $a\sin(u)+b\cos(u)=\sqrt{a^2+b^2}\sin(u+\sin^{-1}(\frac{b}{\sqrt{a^2+b^2}}))$
and peform the substiution $v=u+... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Can you help me out with this? How to find it? $\\int_{0}^{2\\pi }\\frac{sinu}{acosu+bsinu+c}du$",
"content_html": "Can you help me out with this? How to find it? <img src=\"//latex.artofproblemsolvi... | Compute the integral
\[
\int_{0}^{2\pi} \frac{\sin u}{a\cos u + b\sin u + c}\,du.
\] | [
"/Mathematics/CalculusandAnalysis/Calculus/IntegralCalculus",
"/Mathematics/CalculusandAnalysis/Calculus/Integrals/DefiniteIntegrals"
] | Shift the angle to rewrite a cos u + b sin u as R sin(u+φ) and then apply the tan(u/2) substitution. | 170,555 | [
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aops_2678075 | With $x=t^3$ we find $\int {{t^2 \cdot 3t^2 dt}\over {t^3+1}}=3\int (t-{t\over {t^3+1}})dt=3\int (t+{{1/3}\over {t+1}}-{{1/3(t+1)}\over {t^2-t+1}})dt=3\int (t+{{1/3}\over {t+1}}-{{1/6(2t-1)}\over {t^2-t+1}}-{{1/2}\over {t^2-t+1}})dt$.
For the final term we can use $(t-1/2)^2+3/4$ and we find a version of the $\arctan$ ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Evaluate $$\\int\\frac{x^\\frac{2}{3}}{1+x} dx$$",
"content_html": "Evaluate <img src=\"//latex.artofproblemsolving.com/5/d/4/5d40f6701a7899c41aff6f717df11255c40288d9.png\" class=\"latexcenter\" alt=... | Evaluate
\[
\int \frac{x^{2/3}}{1+x}\,dx.
\] | [
"/Mathematics/CalculusandAnalysis/Calculus/IntegralCalculus",
"/Mathematics/CalculusandAnalysis/Calculus/Integrals/IndefiniteIntegrals"
] | Use the substitution x = t^3 to turn the integral into a rational function. | 211,489 | [
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ours_25592 | The numbers \(9^{a}\) and \(2013^{c}\) are divisible by 3, so the numbers \(2^{b}\) and \(2014^{d}\) must give the same remainder when divided by 3. Note that the number 2014 gives a remainder of 1 when divided by 3, so every power of it also gives a remainder of 1 when divided by 3. Thus, \(2^{b}\) must give a remaind... | null | {
"competition": "serbian_mo",
"dataset": "Ours",
"posts": null,
"source": "bilten2013-2-2.md"
} | Prove that there are no natural numbers \(a, b, c, d\) such that
\[
9^{a} + 2^{b} + 2013^{c} = 2014^{d}
\] | [
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"/Mathematics/NumberTheory/Congruences/... | Apply congruences modulo 3 and 4 to force b even and d=1, then use a size bound to obtain a contradiction. | 68,969 | [
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numina_10206486 | To evaluate the integral \( \int_{-1}^{a^2} \frac{1}{x^2 + a^2} \, dx \) without using \( \tan^{-1} x \) or complex integrals, we can use a substitution method.
1. **Substitution**:
Let \( x = a \tan(\theta) \). Then, \( dx = a \sec^2(\theta) \, d\theta \).
2. **Change of Limits**:
When \( x = -1 \):
\[
-... | \frac{\pi}{2a} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "aops_forum"
} | Evaluate $ \int_{ \minus{} 1}^ {a^2} \frac {1}{x^2 \plus{} a^2}\ dx\ (a > 0).$
You may not use $ \tan ^{ \minus{} 1} x$ or Complex Integral here. | [
"/Mathematics/CalculusandAnalysis/Calculus/IntegralCalculus",
"/Mathematics/CalculusandAnalysis/Calculus/Integrals/DefiniteIntegrals"
] | Use the trig substitution x = a·tanθ to turn the denominator into a²·sec²θ, canceling the derivative and yielding a constant integrand. | 101,911 | [
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numina_10030790 | The solution is as follows. $1 \cdot \check{i} s t e p$. Introduce a new unknown function of substitution $t=x+y$, then $\frac{d t}{d x}=1+\frac{1}{t^{2}}$.
$2-\check{u}$ step. Separate the variables: $\frac{t^{2} d t}{1+t^{2}}=d x$, integrate both sides, each with respect to its own variable:
$$
\int \frac{t^{2} d t... | (x+y)-\operatorname{arctg}(x+y)=x+1-\frac{\pi}{4} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 4.3. Solve the Cauchy problem: $\frac{d y}{d x}=\frac{1}{(x+y)^{2}}, y(0)=1$. | [
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"/Mathematics/CalculusandAnalysis/Calculus/DifferentialEquations/OrdinaryDifferentialEquations",
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"/Mathematics/Calculu... | Introduce t = x + y to convert the ODE into a separable equation for t. | 78,844 | [
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ours_17896 | "The given inequality can be transformed into:\n\n\\[\nx^{3}(x+1) + y^{3}(y+1) + z^{3}(z+1) \\geq \\(...TRUNCATED) | null | {
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aops_431847 | "The area of any triangle is one-half the length of any base times the length of the altitude to tha(...TRUNCATED) | null | {"competition":null,"dataset":"AOPS","posts":[{"attachments":[],"content_bbcode":"A right triangle h(...TRUNCATED) | "A right triangle has hypotenuse \\(75\\) cm and one leg \\(45\\) cm. What is the length, in centime(...TRUNCATED) | ["/Mathematics/Geometry/GeneralGeometry/EuclideanGeometry","/Mathematics/Geometry/GeneralGeometry/Ge(...TRUNCATED) | "Equate the triangle's area using its legs with its area using the hypotenuse and altitude, then sol(...TRUNCATED) | 253,285 | [24120,22046,65566,14808,67576,39629,26185,64445,22970,7321,26562,14281,52576,20150,18288,14265,868,(...TRUNCATED) | [0.88525390625,0.9296875,0.96240234375,0.8974609375,0.8623046875,0.86572265625,0.861328125,0.9624023(...TRUNCATED) | [0.0,0.125,0.0,0.0625,0.0625,0.11767578125,0.0999755859375,0.047607421875,0.052642822265625,0.1875,0(...TRUNCATED) | [
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