id string | solution string | answer string | metadata dict | problem string | candidates list | relevance_scores list | relevance_scores_full null |
|---|---|---|---|---|---|---|---|
numina_10111898 | 8. Proof 1: Let $f(x)=\sum_{k=1}^{n}(-1)^{k+1} C_{n}^{k} \frac{1}{k}\left[1-(1-x)^{k}\right]$, then $f(0)=0$, and $f^{\prime}(x)=\sum_{k=1}^{n}(-1)^{k+1} C_{n}^{k}$.
$$
(1-x)^{k-1}=\frac{1}{1-x}\left[1-\sum_{k=0}^{n}(-1)^{n} C_{n}^{k}(1-x)^{k}\right]=\frac{1}{1-x}\left\{1-[1-(1-x)]^{n}\right\}=\frac{1-x^{n}}{1-x}=1+x+\... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 8. Prove: $\sum_{k=1}^{n}(-1)^{k+1} C_{n}^{k} \cdot \frac{1}{k}\left[1-(1-x)^{k}\right]=x+\frac{x^{2}}{2}+\cdots+\frac{x^{n}}{n}(x \in \mathbf{R})$. | [
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aops_46779 | [quote="Slizzel"]If $M$ is the coeficent of $x^2$ from the following exeresion $E(x)=(1+x)^3+(1+x)^4+...+(1+x)^{11}$, find $M$.
p.s. [i]can anybody give some methods in order to solve this kind of problems.[/i][/quote]
i think it should be 3C2 +4C2 +5C2 +...+11C2 :P | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "If $M$ is the coeficent of $x^2$ from the following exeresion $E(x)=(1+x)^3+(1+x)^4+...+(1+x)^{11}$, find $M$. \r\n\r\np.s. [i]can anybody give some methods in order to solve this kind of problems.[/i]",
... | If \(M\) is the coefficient of \(x^2\) in the expression
\[
E(x)=(1+x)^3+(1+x)^4+\cdots+(1+x)^{11},
\]
find \(M\). | [
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53... | null | null |
numina_10136627 | 【Solution】Solution: According to the problem, we know:
Since the 5th number is 53, it indicates that the 7th number is 53 more than the 6th number, and the sum of the 7th and 6th numbers is 225.
Therefore, according to the sum and difference formula, the 6th number is: $(225-53) \div 2=86$.
So the 4th number is $86-53... | 7 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 6. (10 points) Arrange 8 numbers in a row from left to right. Starting from the third number, each number is exactly the sum of the two numbers before it. If the 5th number and the 8th number are 53 and 225, respectively, then the 1st number is $\qquad$ . | [
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numina_10092069 | 6. $\frac{1}{n(n+1)}$.
When $n \geqslant 2$, since $a_{1}+a_{2}+\cdots+a_{n}=n^{2} a_{n}, a_{1}+a_{2}+\cdots+a_{n-1}=(n-1)^{2} a_{n-1}$, we have $a_{n}=n^{2} a_{n}-(n-1)^{2} a_{n-1}$.
From this, we get $a_{n}=\frac{n-1}{n+1} a_{n-1}, n=2,3, \cdots$
Thus, we obtain $\quad a_{n}=\frac{1}{n(n+1)}, n=1,2,3, \cdots$ | \frac{1}{n(n+1)} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 6. The sequence $a_{1}, a_{2}, a_{3}, \cdots$ satisfies (1) $a_{1}=\frac{1}{2}$, (2) $a_{1}+a_{2}+\cdots+a_{n}=n^{2} a_{n}(n \geqslant 2)$, then $a_{n}=$ $\qquad$ | [
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numina_10179111 | 2. The given relation is written as:
$$
\begin{aligned}
& 3\left(\frac{1}{1}-\frac{1}{2}\right)+4\left(\frac{1}{2}-\frac{1}{3}\right)+5\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+(n+2)\left(\frac{1}{n}-\frac{1}{n+1}\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{n}\right) \\
& \frac{3}{1}-\frac{3}{2}... | 2-\frac{1}{n+1} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 2. Demonstrate that $\frac{3}{1 \cdot 2}+\frac{4}{2 \cdot 3}+\frac{5}{3 \cdot 4}+\ldots+\frac{n+2}{n \cdot(n+1)}-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{n}\right)<2, \forall n \in \mathbb{N}, n \geq 2$.
RMCS nr. 29/2009 | [
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numina_10098897 | $\begin{array}{lllll}59 & \end{array}$
$$
\begin{aligned}
x^{n+1}-\sum_{k=0}^{n-1}(n-k) C_{n+1}^{k} x^{k} & =x^{n+1}-\sum_{k=0}^{n-1}(n+1-k) C_{n+1}^{k} x^{k}+\sum_{k=0}^{n-1} C_{n+1}^{k} x^{k} \\
& =\sum_{k=0}^{n+1} C_{n+1}^{k} x^{k}-C_{n+1}^{n} x^{n}-\sum_{k=0}^{n-1}(n+1) C_{n}^{k} x^{k} \\
& =(1+x)^{n+1}-(n+1)(1+x)^... | x\geqslantn | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 59 Given that $n$ is a natural number. If
$$
x^{n+1} \geqslant \sum_{k=0}^{n-1}(n-k) C_{n+1}^{k} x^{k}
$$
Find the range of values for $x$. | [
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aops_2834582 | Note that $(n+1)x_{n+1}-nx_n=x_n^2$, so summing gives $(n+1)x_{n+1}=2+x_1^2+...+x_n^2>\sum x_i^2 \geq \frac {(\sum x_i)^2}{n} $, done. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $(x_n)_{n=1}^\\infty$ be a sequence defined recursively with: $x_1=2$ and $x_{n+1}=\\frac{x_n(x_n+n)}{n+1}$ for all $n \\ge 1$. Prove that $$n(n+1) >\\frac{(x_1+x_2+ \\ldots +x_n)^2}{x_{n+1}}.$$\n\n[i]... | Let \((x_n)_{n=1}^\infty\) be a sequence defined recursively by
\[
x_1=2,\qquad x_{n+1}=\frac{x_n(x_n+n)}{n+1}\quad\text{for all }n\ge1.
\]
Prove that
\[
n(n+1)>\frac{(x_1+x_2+\cdots+x_n)^2}{x_{n+1}}.
\] | [
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numina_10130920 | Let \( S = \sum_{k=0}^{\left[\frac{n}{2}\right]}(-1)^{k} C_{n}^{k} C_{2 n-2 k-1}^{n-1} = \sum_{k=0}^{\left[\frac{n}{2}\right]}(-1)^{k} C_{n}^{k} C_{2 n-2 k-1}^{n-2 k} \), then \( S \) is the coefficient of \( x^n \) in \( \sum_{k=0}^{\left[\frac{n}{2}\right]}(-1)^{k} C_{n}^{k} x^{2 k} (1+x)^{2 n-2 k-1} \). The term \( ... | 1 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 4 Proof: $\sum_{k=0}^{\left[\frac{n}{2}\right]}(-1)^{k} C_{n}^{k} C_{2 n-2 k-1}^{n-1}=1(n \geqslant 1)$. | [
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ours_12303 | The sum telescopes as
$$
\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)+\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)+\cdots+\left(\frac{1}{14^{2}}-\frac{1}{15^{2}}\right)=\frac{1}{1^{2}}-\frac{1}{15^{2}}=\frac{224}{225}
$$
Thus, the value of the sum is \(\frac{224}{225}\). Therefore, the answer is $224 + 225 = \b... | 449 | {
"competition": "hmmt",
"dataset": "Ours",
"posts": null,
"source": "alg_feb_2002.md"
} | Determine the value of the sum
$$
\frac{3}{1^{2} \cdot 2^{2}}+\frac{5}{2^{2} \cdot 3^{2}}+\frac{7}{3^{2} \cdot 4^{2}}+\cdots+\frac{29}{14^{2} \cdot 15^{2}}
$$ If the answer is of the form of an irreducible fraction $\frac{a}{b}$, compute the value of $a + b$. | [
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aops_1594911 | You are finding the number of $(a,c,b',d')$ because that determines $(a,b,c,d)$ uniquely and that determines the polynomial uniquely. From there on, it is a straight application to of stars of bars to $a+c+b'+d'=9$ to find the number of $(a,c,b',d')$. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "2018 AMC 12B Problem 22\n\n[b]QUESTION : I don't get why it is 12C3 in the solution. How would stars and bars work in this problem? Could someone please clarify. : [/b]\n\n\nProblem :\nConsider polynomials... | Consider polynomials \(P(x)\) of degree at most \(3\), each of whose coefficients is an element of \(\{0,1,2,3,4,5,6,7,8,9\}\). How many such polynomials satisfy \(P(-1)=-9\)?
(A) \(110\) (B) \(143\) (C) \(165\) (D) \(220\) (E) \(286\)
Suppose \(P(x)=ax^3+bx^2+cx+d\). Then \(P(-1)=-a+b-c+d=-9\), so \(-a+b-c+d=-9\... | [
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aops_1852880 | [quote=TuZo]A possible solution for Q2:
Using Newton binomial theoreme we have: ${{(1+x)}^{n}}=\sum\limits_{k=0}^{n}{C_{n}^{k}{{x}^{k}}}$. Deriving both side, we get: $n{{(1+x)}^{n-1}}=\sum\limits_{k=1}^{k}{nC_{n}^{k}{{x}^{k-1}}}\Rightarrow \sum\limits_{k=1}^{n}{kC_{n}^{k}=n\cdot {{2}^{n-1}}}$[/quote]
thanks ..... :la... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Q1 Find the number of $30-$digit combinations (where repetition is allowed) from the set {$0, 1, 2, 3, 4, 5$} in\nwhich each digit $i$ in the set occurs at least $i$ times in the combination.eg....if digi... | Q1. Find the number of 30-digit sequences (repetition allowed) from the set \(\{0,1,2,3,4,5\}\) in which each digit \(i\) in the set occurs at least \(i\) times in the sequence.
Q2. Evaluate
\[
\binom{100}{1}+2\binom{100}{2}+3\binom{100}{3}+\cdots+100\binom{100}{100}.
\] | [
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aops_499178 | $\sum_{n=100}^m\binom{n}{100}= $ the coefficient of $ x^{100} $ in $(1+x)^{100} + (1+x)^{101} +...+(1+x)^m $ = $\binom{m+1}{101} $
So, $\sum_{m=100}^{201}\sum_{n=100}^{m}\binom{201}{m+1}\binom{n}{100}$ = $\sum_{m=100}^{201}\binom{201}{m+1}\binom{m+1}{101} $ = $\frac{201!}{101! 100!}\sum_{m=100}^{201}\binom{100}{m-10... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find \\[ \\sum_{m=100}^{201}\\sum_{n=100}^{m}\\binom{201}{m+1}\\binom{n}{100} \\]",
"content_html": "Find <img src=\"//latex.artofproblemsolving.com/7/b/1/7b1f1907fac6a4b28937249d1e8afe5b802a6817.png... | Find
\[
\sum_{m=100}^{201}\sum_{n=100}^{m}\binom{201}{m+1}\binom{n}{100}.
\] | [
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aops_3384604 | [hide=Solution for the fourth one]
Note that $$\binom{2n-1}{k} = \binom{2n-1}{2n-1-k},$$ for all $k \in \{0, 1, \ldots, n-1\},$ and we also have $$\sum_{k=0}^{2n-1} \binom{2n-1}{k} = 2^{2n-1}$$ by the Binomial Theorem, so it follows that
\begin{align*}
\sum_{k=0}^{n-1} \binom{2n-1}{k} &= \frac{1}{2} \cdot \sum_{k=0}^{n... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Show that:$\\sum_{k=0}^{n} \\frac{(-1)^{k} }{(k+1)^{2} } \\binom{n}{k} =\\frac{1}{n+1} \\sum_{k=0}^{n} \\frac{1}{k+1}$ \nShow that:$\\sum_{k=0}^{n} (-1)^{k} \\binom{n}{k} ^{-1} =\\frac{n+1}{n+2} (1+(-1)^{n... | 1) Show that
\[
\sum_{k=0}^{n} \frac{(-1)^{k}}{(k+1)^{2}}\binom{n}{k}=\frac{1}{n+1}\sum_{k=0}^{n}\frac{1}{k+1}.
\]
2) Show that
\[
\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}^{-1}=\frac{n+1}{n+2}\bigl(1+(-1)^{n}\bigr).
\]
3) Show that
\[
\sum_{k=0}^{n}k\binom{2n}{k}=n2^{2n-1}.
\]
4) Show that
\[
\sum_{k=0}^{n-1}\binom{2n-1}{... | [
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aops_2056128 | [quote=awesomeguy856][hide=complementary counting] Total ways are 15 choose 6, which is 5005. the number of ways to go through P is (6 choose 3)*(9 choose 3)=1680, and the number of ways to go through Q is (10 choose 4)*(5 choose 2)=2100 so it is 5005-1680-2100=1225.[/quote]
We must add back the number of paths that g... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "A person starts from the origin in the X-Y plane. He takes steps of one unit along the positive x axis or positive y-axis. Travelling in this manner find the total number of ways he can reach A (9,6) avoid... | A person starts from the origin in the XY-plane. He takes steps of one unit along the positive \(x\)-axis or positive \(y\)-axis. Travelling in this manner, find the total number of ways he can reach \(A(9,6)\) avoiding both the points \(P(3,3)\) and \(Q(6,4)\). | [
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numina_10190454 | Solution. We will prove the statement by induction on $n$. Clearly, $0<f(1)<\sqrt{3}$. Assume that the statement holds for $n=k$, i.e., $0<f(k)<\sqrt{3}$. We have, $f(k+1)=2-\frac{1}{f(k)+2}$. Therefore, considering that $0<f(k)<\sqrt{3}$, we get that $0<2-\frac{1}{2}<f(k+1)<2-\frac{1}{\sqrt{3}+2}=\sqrt{3}$, i.e., the ... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 4A. For every natural number $n$, it holds that $f(n+1)=\frac{2 f(n)+3}{f(n)+2}$ and $f(0)=\frac{1}{4}$. Prove that for every natural number $n$, it holds that $0<f(n)<\sqrt{3}$. | [
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numina_10071232 | It is easy to check the equality $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$.
Applying this equality for k from 1 to 2002, we write
$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{2002 \cdot 2003}=$
$=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\fra... | \frac{2002}{2003} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Find the sum. Find the sum
$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{2002 \cdot 2003}$ | [
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3468,
... | null | null |
numina_10147181 | 9. $a_{n}=\frac{3}{n}$. Let $b_{n}=n a_{n}$, therefore $b_{n+1}=b_{n}$, so $b_{n}=b_{1}=a_{1}=3$. Hence $a_{n}=\frac{3}{n}$. | a_{n}=\frac{3}{n} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 9. In the sequence $\left\{a_{n}\right\}$, if $a_{1}=3$, and $n a_{n}=(n+1) a_{n+1}$, then $a_{n}=$ | [
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numina_10182550 | The binomial theorem states
$$
(a+b)^{n}=\sum_{k=0}^{n}\binom{n}{k} a^{n-k} b^{k}
$$
Specifically, it follows from this for $a=1, b=1$:
$$
(1+1)^{n}=\sum_{k=0}^{n}\binom{n}{k} 1^{n-k} 1^{k}=\sum_{k=0}^{n}\binom{n}{k}=2^{n}
$$
and for $a=1, b=3$:
$$
(1+3)^{n}=\sum_{k=0}^{n}\binom{n}{k} 1^{n-k} 3^{k}=\sum_{k=0}^{n}\... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | ## Task 27/78
It is to be shown that for every natural number $n$ the following holds
$$
\left[\sum_{k=0}^{n}\binom{n}{k}\right]^{2}=\sum_{k=0}^{n}\binom{n}{k} \cdot 3^{k}
$$ | [
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... | null | null |
aops_545509 | [hide="Sol"]
We use the method of [b] Partial Fraction Decomposition [/b]. This gives us in general \[\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\]. Therefore \[\sum_{i=1}^{20}(\frac{1}{i(i+1)})=\sum_{i=1}^{20}(\frac{1}{i}-\frac{1}{i+1})=1-\frac{1}{21}=\frac{20}{21}\]
Henceforth $a+b=\boxed{41}$.
[/hide]
[hide="I... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "The sum $T=\\frac{1}{1*2}+\\frac{1}{2*3}+...+\\frac{1}{20*21}$ can be written in the form $\\frac{a}{b}$ where $a$ and $b$ are relatively prime positive integers. The value of $a+b$ is?",
"content_ht... | The sum
\[
T=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots+\frac{1}{20\cdot 21}
\]
can be written in the form \(\dfrac{a}{b}\) where \(a\) and \(b\) are relatively prime positive integers. Find \(a+b\). | [
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55692,... | null | null |
numina_10146820 | A direct proof of the above equation is easy, according to the binomial theorem:
$$
(x+y)^{n}=\binom{n}{0} x^{n}+\binom{n}{1} x^{n}{ }^{1} y+\binom{n}{2} x^{n-2} y^{2}+\cdots+\binom{n}{n} y^{n}
$$
Let $x=1, y=-1$ be substituted, and the equation is proved.
Below is an explanation with combinatorial meaning:
The signif... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 1-36 $\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\cdots \pm\binom{ n}{n}=0$. | [
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aops_535648 | [quote="hoangquan"]Let $1 \leq j\leq i$ are integer. Using Lagrange interpolation ,prove that:
\[ \sum\limits_{p=0}^{j} (-1)^pC_i^pC_{i+j-p-1}^{j-p}=0\][/quote]
I think you should apply Lagrange interpolation into polynomial $P(x)=(i+1-x)(i+2-x)...(i+j-1-x)$ with chosen points $x_{k}=k \quad (k=\overline{0;j})$.
====... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $1 \\leq j\\leq i$ are integer. Using Lagrange interpolation ,prove that:\n\\[ \\sum\\limits_{p=0}^{j} (-1)^pC_i^pC_{i+j-p-1}^{j-p}=0\\]",
"content_html": "Let <img src=\"//latex.artofproblemsol... | Let \(1\le j\le i\) be integers. Using Lagrange interpolation, prove that
\[
\sum_{p=0}^{j} (-1)^p \binom{i}{p}\binom{i+j-p-1}{\,j-p\,}=0.
\] | [
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aops_1883416 | $(1- \frac{1}{2} )^n=\sum\limits_{k=0}^{n}(-1)^k \frac{\binom{n}{k}}{2^k}$
I guess so ,though i forget a lot. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Evaluate $$\\sum_{k=0}^{n} (-1)^k \\frac{\\binom{n}{k}}{2^k}$$",
"content_html": "Evaluate <img src=\"//latex.artofproblemsolving.com/e/5/8/e58c62c5464a591bd892e9b53ef31c51097bee40.png\" class=\"late... | Evaluate
\[
\sum_{k=0}^{n} (-1)^k \frac{\binom{n}{k}}{2^k}.
\] | [
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numina_10113828 | 3. $\frac{4037}{2019}$.
$$
\begin{array}{l}
\text { Given } f(n)-f(n-1) \\
=\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n},
\end{array}
$$
we know
$f(2)-f(1)=1-\frac{1}{2}$,
$f(3)-f(2)=\frac{1}{2}-\frac{1}{3}$,
$\qquad$
$f(2019)-f(2018)=\frac{1}{2018}-\frac{1}{2019}$.
Adding the above equations on both sides, we get
$$
\b... | \frac{4037}{2019} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 3. The function $f(x)$ satisfies $f(1)=1$, and
$$
f(n)=f(n-1)+\frac{1}{n(n-1)}\left(n \geqslant 2, n \in \mathbf{Z}_{+}\right) \text {. }
$$
Then $f(2019)=$ | [
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aops_497162 | [hide]The sum is the coefficient of $x^m$ in $(x+2x^2+3x^3+4x^4+...)^n$.
\begin{align*}x+2x^2+3x^3+4x^4+...&=x\cdot\frac{\mathrm{d}}{\mathrm{d}x}(1+x+x^2+x^3+...)\\&=x\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{1-x}\right)\\&=x\cdot \frac{1}{(1-x)^2}\end{align*}
So the expression is $\left(\frac{x}{(1-x)^2}\rig... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Prove that $\\sum_{(a_1,a_2,...,a_n)\\in S} a_1\\cdot a_2\\cdot ... \\cdot a_n = \\binom{m+n-1}{2n-1}$ where S is the set of all $(a_1, a_2,..., a_n)$ such that $a_1 + a_2 + ... + a_n = m$ and $a_i$ are po... | Prove that
\[
\sum_{(a_1,a_2,\dots,a_n)\in S} a_1a_2\cdots a_n = \binom{m+n-1}{2n-1},
\]
where \(S=\{(a_1,a_2,\dots,a_n)\in\mathbb{Z}^n: a_i\ge1,\ a_1+\cdots+a_n=m\}\). | [
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numina_10147286 | 6. $\frac{2^{n+1}}{n+1}-2$.
From $S_{n}^{2}+4 S_{n}-3 T_{n}=0$,
then $S_{n+1}^{2}+4 S_{n+1}-3 T_{n+1}=0$.
(2) - (1) gives
$$
\left(S_{n+1}+S_{n}\right) a_{n+1}+4 a_{n+1}-3 a_{n+1}^{2}=0 \text {. }
$$
Since $a_{n+1} \neq 0$, then
$$
\left(S_{n+1}+S_{n}\right)+4-3 a_{n+1}=0 \text {. }
$$
$$
\text { Hence }\left(S_{n}+S... | \frac{2^{n+1}}{n+1}-2 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 6. Given a sequence $\left\{a_{n}\right\}$ whose terms are all non-zero. Let the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ be $S_{n}$, and the sum of the first $n$ terms of the sequence $\left\{a_{n}^{2}\right\}$ be $T_{n}$, and $S_{n}^{2}+4 S_{n}-3 T_{n}=0\left(n \in \mathbf{Z}_{+}\right)$. The... | [
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aops_329562 | It is easily proven by induction (and in fact can be generalized) that $ 1 \plus{} 2^1 \plus{} 2^2 \plus{} 2^3\dots \plus{} 2^{n \minus{} 1} \plus{} 2^{n} \equal{} 2^{n \plus{} 1} \minus{} 1$.
Thus, $ 2^{10} \minus{} 1 \equal{} \boxed{1023}$. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$ 2^0\\plus{}2^1\\plus{}2^2\\plus{}2^3\\plus{}2^4\\plus{}2^5\\plus{}2^6\\plus{}2^7\\plus{}2^8\\plus{}2^9\\equal{}\\boxed{?}$",
"content_html": "<img src=\"//latex.artofproblemsolving.com/1/2/c/12c229... | \(2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + 2^9 = \boxed{?}\) | [
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numina_10110882 | Let \( S_{n, r} = \sum_{k=0}^{n-r} (-2)^{-k} C_{n}^{r+k} C_{n+r+k}^{k} \). By substituting \( k \) with \( n-r-k \), we get
\[
S_{n, r} = \sum_{k=0}^{n-r} (-2)^{-n+r+k} C_{n}^{n-k} C_{2n-k}^{n-r-k} \text{. Thus, } S_{n, r} \text{ is the coefficient of } x^{n-r} \text{ in the expansion of } \sum_{k=0}^{n-r} (-2)^{-n+r+k... | 2^{r-n}(-1)^{\frac{n-r}{2}}C_{n}^{\frac{n-r}{2}} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 3 Let $n, r$ be positive integers, $r < n$ and $n-r$ be even. Prove:
$$
\sum_{k=0}^{n-r}(-2)^{-k} C_{n}^{r+k} C_{n+r+k}^{k}=2^{r-n}(-1)^{\frac{n-r}{2}} C_{n}^{\frac{n-r}{2}} .
$$ | [
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aops_1659649 | This is a telescoping sum: $\sum\limits_{k=1}^{n}{\left( k+\frac{1}{3k-2}-\frac{1}{3k+1} \right)=\frac{n(n+1)}{2}+1-\frac{1}{3n+1}}$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "find sum of first $n$ terms of sequence\n$\\frac{7}{1 * 4} +\\frac{59}{4*7}+\\frac{213}{7*10}+....+\\frac{9n^3-3n^2-2n+3}{(3n-2)(3n+1)}$",
"content_html": "find sum of first <img src=\"//latex.artofp... | Find the sum of the first \(n\) terms of the sequence
\[
\frac{7}{1\cdot 4} + \frac{59}{4\cdot 7} + \frac{213}{7\cdot 10} + \cdots + \frac{9n^3-3n^2-2n+3}{(3n-2)(3n+1)}.
\] | [
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ours_27931 | Put \(p_{n} = a_{n} - a_{n-1}\). Then the statement of the problem implies the formula \(p_{n} = p_{n-1} + 1\), showing that the numbers \(p_{n}\) form an arithmetic progression with unity as the common difference. Therefore, \(p_{n} = p_{2} + n - 2\). Now we find
\[
\begin{aligned}
a_{n} &= (a_{n} - a_{n-1}) + (a_{n-1... | null | {
"competition": "misc",
"dataset": "Ours",
"posts": null,
"source": "Problems in Elementary Mathematics - group_24.md"
} | It is known that the number sequence \(a_{1}, a_{2}, a_{3}, \ldots\) satisfies, for any \(n\), the relation
\[ a_{n+1} - 2a_{n} + a_{n-1} = 1 \]
Express \(a_{n}\) in terms of \(a_{1}, a_{2}\), and \(n\). | [
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aops_27539 | Bleh - I'm lazy so easy way:
[hide]$\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3} + \ldots + \frac{1}{256}-\frac{1}{257} = 1 - \frac{1}{257} = \frac{256}{257}$
But ofcourse, this appears in a millisecond to the trained telescoping eye.[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Here is a classic problem that appeared in many past math competitions.\r\n\r\nBut in this problem, I want you to find 3 different ways to do this.\r\n\r\nPost as many ways as possible you can think of. I ... | Calculate
\[
\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots+\frac{1}{256\cdot 257}.
\] | [
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3... | null | null |
numina_10136370 | 8. $\left(q_{m+n}\right.$ Consider $(1+x)^{n}$ and $(1+x)^{m}$ these two generating functions, on one hand $(1+x)^{n}(1+x)^{m}=(1+x)^{m+n}$ the coefficient of $x^{q}$ is $C_{m+n}$; on the other hand, the coefficient of $x^{q}$ is $\sum_{k=0}^{\eta} C_{n}^{k} C_{m}^{q-k}$. | C_{+n}^{q} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 8. Simplify $\sum_{k=0}^{4} C_{n}^{k} C_{m}^{q-k}=$ $\qquad$ (Vandermonde's formula) | [
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aops_436911 | ... and of course $n > 1$; for $n=1$ you get $-1 \neq 0$; this is because the derivative of $(1-x)^1$, computed at $x=1$, does not vanish ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $n$ be a positive integer greater than 1.\nProve that : \n\n$\\sum_{k=1}^{n}(-1)^{k} \\cdot k \\cdot \\binom{n}{k}=0$",
"content_html": "Let <img src=\"//latex.artofproblemsolving.com/1/7/4/174fa... | Let \(n\) be a positive integer greater than 1. Prove that
\[
\sum_{k=1}^{n}(-1)^{k} k \binom{n}{k} = 0.
\] | [
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numina_10080873 | We recognize a binomial of Newton for which $a=1$ and $b=1$. The answer is therefore $2^{n}$. | 2^n | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | $$
\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\ldots+\binom{n}{k}+\ldots+\binom{n}{n}=?
$$ | [
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42859,
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... | null | null |
numina_10119032 | Prove that from the identity $k C_{n}^{k}=n C_{n-1}^{k-1}$, we get
$$
\sum_{k=1}^{n} k C_{n-1}^{k-1}=\sum_{k=1}^{n} n C_{n-1}^{k-1}=n \sum_{k=0}^{n-1} C_{n-1}^{k}=n \cdot 2^{n-1} .
$$ | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 1 Proof: $\sum_{k=1}^{n} k C_{n}^{k}=n \cdot 2^{n-1}$. | [
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numina_10057476 | Each sum can be represented as a binomial expansion
a) $(1+2)^{5}$;
b) $(1-1)^{n}$;
c) $(1+1)^{n}$.
a) $3^{5}$;
b) 0;
c) $2^{n}$. | 3^5,0,2^n | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | [Pascal's Triangle and Newton's Binomial]
Calculate the sums:
a) $C_{5}^{0}+2 C_{5}^{1}+2^{2} C_{5}^{2}+\ldots+2^{5} C_{5}^{5}$
b) $C_{n}^{0}-C_{n}^{1}+\ldots+(-1)^{n} C_{n}^{n}$
c) $C_{n}^{0}+C_{n}^{1}+\ldots+C_{n}^{n}$. | [
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numina_10100748 | 10. (1) (Method 1) It is easy to find that $a_{1}=1, a_{2}=\frac{3}{2}, a_{3}=\frac{7}{4}, a_{4}=\frac{15}{8}$. From this, we conjecture: $a_{n}=\frac{2^{n}-1}{2^{n-1}}$.
We will prove this by mathematical induction.
When $n=1$, $a_{1}=S_{1}=2-a_{1} \Rightarrow a_{1}=1$, the conclusion holds.
Assume that when $n=k\lef... | \frac{5}{3} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 10. (20 points) Let the sequence $\left\{a_{n}\right\}$ have the sum of the first $n$ terms as $S_{n}$, and $S_{n}=2 n-a_{n}\left(n \in \mathbf{N}^{*}\right)$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) If the sequence $\left\{b_{n}\right\}$ satisfies $b_{n}=2^{n-1} a_{n}$, prove that... | [
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numina_10097726 | 3. Since $\frac{1}{k+1} C_{n}^{k}=\frac{1}{n+1} C_{n+1}^{k+1}$, therefore, the original expression $=\frac{2^{n+1}-1}{n+1}$. | \frac{2^{n+1}-1}{n+1} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 3. Prove: $C_{n}^{0}+\frac{1}{2} C_{n}^{1}+\frac{1}{3} C_{n}^{2}+\cdots+\frac{1}{n+1} C_{n}^{n}=\frac{2^{n+1}-1}{n+1}$. | [
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aops_109310 | [hide="Solution"]\begin{eqnarray*}\sum_{k=1}^{n}a_{k}=n^{2}+1 &\implies& a_{n}=\left(\sum_{k=1}^{n}a_{k}\right)-\left(\sum_{k=1}^{n-1}a_{k}\right)=n^{2}+1-[(n-1)^{2}+1]\\ &\iff& a_{n}=2n-1\end{eqnarray*}[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find the $n$ th term of the sequence $\\{a_{n}\\}$ such that $\\frac{a_{1}+a_{2}+\\cdots+a_{n}}{n}=n+\\frac{1}{n}\\ \\ (n=1,\\ 2,\\ 3,\\ \\cdots).$",
"content_html": "Find the <img src=\"//latex.art... | Find the \(n\)th term of the sequence \(\{a_n\}\) such that
\[
\frac{a_1+a_2+\cdots+a_n}{n}=n+\frac{1}{n}\qquad(n=1,2,3,\ldots).
\] | [
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aops_81366 | It is similar to British Mathematical Olympiad, 1996, question 2.
[hide]
$f(1)+f(2)=2^2f(1)$ implies $f(2)=\frac{1}{(2^2-1)}\cdot f(1)$.
Next,
$f(3)=\frac{1}{3^2-1}\cdot (f(1)+f(2))=\frac{1}{3^2-1}\cdot [ f(1) + \frac{1}{2^2-1}] \cdot f(1) \\ = \frac{1}{2^2-1}\cdot \frac{2^2}{3^2-1} \cdot f(1)$
$f(4) = ........$
Good ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let the function $f: \\mathbb{R}\\to\\mathbb{R}$ satisfy:\r\n$f(1)=2007$ and $f(1)+f(2)+\\cdots+f(n)=n^2f(n), \\forall n \\in \\mathbb{N}$.\r\nFind $f(2006)$.",
"content_html": "Let the function <img... | Let \(f:\mathbb{R}\to\mathbb{R}\) satisfy
\[
f(1)=2007\quad\text{and}\quad f(1)+f(2)+\cdots+f(n)=n^2f(n)\quad\forall n\in\mathbb{N}.
\]
Find \(f(2006)\). | [
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aops_3123421 | Note that $a_n = \left(\sqrt{1 + \sqrt{a_{n-1}}} - 1\right)^2$ for all $n \ge 2$. Let $b_n = 1 + \sqrt{a_n}$ for all $n \ge 1$, then $b_1 = 2$ and $b_n = \sqrt{b_{n-1}}$ for all $n \ge 2$. So $b_n = 2^{\frac{1}{2^{n-1}}}$ for all $n \ge 1$. Then
\begin{align*}
S_{2023} &= \sum_{k=1}^{2023} 2^ka_k = \sum_{k=1}^{2023} 2... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "The positive sequence $\\{a_n\\}$ satisfies:$a_1=1$ and $$a_n=2+\\sqrt{a_{n-1}}-2 \\sqrt{1+\\sqrt{a_{n-1}}}(n\\geq 2)$$\nLet $S_n=\\sum\\limits_{k=1}^{n}{2^ka_k}$. Find the value of $S_{2023}$.",
"co... | The positive sequence \(\{a_n\}\) satisfies \(a_1=1\) and
\[
a_n = 2 + \sqrt{a_{n-1}} - 2\sqrt{1+\sqrt{a_{n-1}}}\quad (n\ge 2).
\]
Let \(S_n=\sum_{k=1}^{n}2^k a_k\). Find the value of \(S_{2023}\). | [
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... | null | null |
aops_1746332 | If that's what you mean, then this is the solution.
[hide=solution]Using the binomial theorem: $(1+1)^k=\sum_{r=0}^k{k\choose r}1^r\cdot1^{k-r}=2^k$.[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Prove or Disprove $$\\sum_{r=0}^{n} \\binom{n}{r}=2^n$$ For some positive integer $k$",
"content_html": "Prove or Disprove <img src=\"//latex.artofproblemsolving.com/9/e/e/9ee37df46b2486fd9fa7f07acad... | Prove or disprove
\[
\sum_{r=0}^{n} \binom{n}{r} = 2^n
\]
for some positive integer \(n\). | [
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numina_10009936 | Answer. $a_{n}=(n+1) 2^{n-2}$
Solution. Let's find the first few terms of the sequence: $a_{1}=1, a_{2}=\frac{3}{1} \cdot 1=3 \cdot 1$, $a_{3}=\frac{4}{2} \cdot(1+3)=8=4 \cdot 2, a_{4}=\frac{5}{3} \cdot(1+3+8)=20=5 \cdot 4$ and make the assumption that $a_{n}=(n+1) 2^{n-2}$. We also calculate the sums of the first ter... | a_{n}=(n+1)2^{n-2} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 11.4. The sequence of natural numbers $a_{n}, n=1,2, \ldots$ is such that $a_{1}=1$ and $a_{n}=\frac{n+1}{n-1}\left(a_{1}+a_{2}+\ldots+a_{n-1}\right)$ for all $n=2,3, \ldots$. Find the formula for the "general term" of the sequence, that is, a formula that explicitly expresses $a_{n}$ in terms of $n$ for any $n$. | [
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aops_201975 | [quote="Mathpro"]Let $ \{a_n \} _{n \equal{} 1}^{\infty}$ be sequence of real numbers such that $ a_1 \equal{} 2$ and $ a_{n \plus{} 1} \equal{} {a_n}^2 \minus{} {a_n} \plus{} 1$, for $ n \equal{} 1,2,...$ Prove that
$ 1 \minus{} \frac {1}{2005^{2005}} < \frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} \frac {1}{a_3} \pl... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $ \\{a_n \\} _{n \\equal{} 1}^{\\infty}$ be sequence of real numbers such that $ a_1 \\equal{} 2$ and $ a_{n \\plus{} 1} \\equal{} {a_n}^2 \\minus{} {a_n} \\plus{} 1$, for $ n \\equal{} 1,2,...$ Prove ... | Let \(\{a_n\}_{n=1}^{\infty}\) be a sequence of real numbers such that \(a_1=2\) and
\[
a_{n+1}=a_n^2-a_n+1\quad\text{for }n=1,2,\dots
\]
Prove that
\[
1-\frac{1}{2005^{2005}}<\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_{2005}}<1.
\] | [
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aops_1069857 | [hide = Solution]
$\frac{1}{n*(n+1)}$ = $\frac{1}{n}$ - $\frac{1}{(n+1)}$
Using this the sum is 1 - $\frac{1}{2}$ + $\frac{1}{2}$ - $\frac{1}{3}$ +$\frac{1}{3}$ -...+$\frac{1}{19}$ - $\frac{1}{20}$
The terms in between cancel out so the sum is 1 - $\frac{1}{20}$ = $\frac{19}{20}$
$\frac{19}{20}$
[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": " $\\frac{1}{1*2}+\\frac{1}{2*3}+\\frac{1}{3*4}+...\\frac{1}{19*20}=?$\n\nWhat is the sum of the finite series above?",
"content_html": "<img src=\"//latex.artofproblemsolving.com/6/9/f/69f0e7c479e09b... | Clean up the problem statement: remove all comments, side thoughts, indices, hints, or any content not part of the problem itself. Reformat the remaining text using proper LaTeX, do not change the contents of the problem.
\[
\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots+\frac{1}{19\cdot 20} = ?
\]
W... | [
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aops_269552 | [hide=" Solution"]Answer is 4.
Let n^2 - 440 = a^2 where n and a are positive integers.
Rearranging:
n^2 - a^2 = 440
Factoring:
(n + a) (n - a) = 440 = 2 x 2 x 2 x 5 x 11
n and a must be of the same parity, so assign variables:
n + a = 2k
n - a = 2x
Note that n = k + x.
Substituting:
4kx = 440
kx = 5 x 11 x 2
Lo... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "[b]1st PROBLEM[/b]\r\nFor how many natural numbers value of n ; $ n^2 \\minus{} 440$ is a perfect square?\r\n\r\n\r\n$ A) 1 \\hspace{1cm}B) 2\\hspace{1cm }C) 3 \\hspace{1cm}D) 4\\hspace{1cm}E) none \\hspac... | 1st PROBLEM
For how many natural numbers \(n\) is \(n^2 - 440\) a perfect square?
A) 1 \quad B) 2 \quad C) 3 \quad D) 4 \quad E) none of them | [
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aops_10217 | How many ordered triples (x,y,z) of positive integers satisfy xyz + xy + xz + yz + x + y +z = 1000?
xyz + xy + xz + yz + x + y + z + 1 = 1001.
(x+1)(y+1)(z+1) = 1001.
x+1, y+1, and z+1 are factors of 1001 so then factoring 1001, we get a prime factorization of 7*11*13. Since x, y, and z are positive integers,... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How many ordered triples (x,y,z) of positive integers satisfy xyz + xy + xz + yz + x + y +z = 1000?",
"content_html": "How many ordered triples (x,y,z) of positive integers satisfy xyz + xy + xz + yz... | How many ordered triples \((x,y,z)\) of positive integers satisfy
\[
xyz + xy + xz + yz + x + y + z = 1000?
\] | [
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aops_590480 | [hide="Idk"]First we re-arrange to get $mn-m-n=0$. Adding 1 to both sides, we get $mn-m-n+1=1$. Factoring, we get $(m-1)(n-1)=1$. Therefore, $m-1$ and $n-1$ both have to be 1 or both have to be $-1$ for the product to be $1$. Therefore, the pairs that work are (2,2) and (0,0). So the answer is $\mathbf{ B} $[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How many pairs $(m,n)$ of integers satisfy the equation $m+n=mn$?\n\n$\\textbf{(A) }1\\qquad\\textbf{(B) }2\\qquad\\textbf{(C) }3\\qquad\\textbf{(D) }4\\qquad \\textbf{(E) }\\text{more than }4$",
"co... | How many pairs \((m,n)\) of integers satisfy the equation \(m+n=mn\)?
(A) \(1\) \quad (B) \(2\) \quad (C) \(3\) \quad (D) \(4\) \quad (E) more than \(4\) | [
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numina_10045276 | $\triangle$ The methods we used to solve problems 799-802 do not achieve the goal here. Let's try to gather all terms of the equation on the left side and arrange them by the degree of $x$:
$$
x^{2}-(y+1) x+\left(y^{2}-y\right)=0
$$
We will consider the last equation as a quadratic equation in terms of $x$. Its discr... | (1;2),(2;2),(0;0),(1;0),(0;1),(2;1) | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 804. Solve the equation in integers
$$
x^{2}-x y+y^{2}=x+y
$$ | [
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aops_131682 | This is not an open question. Could a moderator please move this?
[b]Solution.[/b] Let $\gcd (x,y)=d$. Then $x=da,\ y=db$.
\[\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\implies z(x+y)=xy.\]
Substituting $x=da,\ y=db$, we have
\[z=\frac{dab}{a+b}\implies d=k(a+b).\]
As a result, $(x,y,z)=(ak(a+b),bk(a+b),kab)\in\mathbb... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Dear friends,\r\ni want 2 know how to solve 2nd degree or more diophantine equations.\r\nI am sending an easy prob\r\n1/x+1/y=1/z\r\nplz help me to solute this nd plz send more probs nd solution\r\n>moon",... | Dear friends,
I want to know how to solve second-degree or higher Diophantine equations.
I am sending an elementary problem:
\[
\frac{1}{x} + \frac{1}{y} = \frac{1}{z}
\]
Please help me to solve this and please send more problems and solutions.
— moon | [
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aops_1678009 | [quote=Mixer_V]We can rewrite the expression as $xy-4y+3x-12=36 \rightarrow (x-4)(y+3)=36=2^2*3^2$
The divisors of $36$ are $2(2+1)(2+1)=18$, hence we have a total of $18$ ordered pairs.
Setting $x-4=a$ and $y+3=b $, with $ab=36$ we can find all the pairs.[/quote]
Subtracting 12 on both sides will give 12 not 36 on ri... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How many integer ordered pairs $(x,y)$ are solutions to the equation $xy-4y+3x=24$?\n\nFollow-up Question: Can you name all the pairs? \n\nSource: I saw this on Facebook",
"content_html": "How many i... | How many integer ordered pairs \((x,y)\) are solutions to the equation
\[
xy - 4y + 3x = 24 ?
\]
Follow-up: Find all such pairs \((x,y)\). | [
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aops_1288007 | [quote=aiyer12]There is a nicer way to express it.
Note that $8n+1$ must be an odd perfect square so let $8n+1=(2k+1)^2=4k^2+4k+1$ which means $2n=k^2+k$. It follows that $m=\frac{2n+1\pm (2k+1)}{2}=\frac{k^2+k\pm (2k+1)}{2}$ which means solutions are of the form $(m,n)=(\frac{k^2+3k+2}{2},\frac{k^2+k}{2})$ and $(m,n)... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find all the integral solutions of $(m-n)^2=m+n$",
"content_html": "Find all the integral solutions of <img src=\"//latex.artofproblemsolving.com/4/7/d/47da2399c7627df7a9d5aaa6e056ad527c5e7cac.png\" ... | Find all integer solutions of \((m-n)^2 = m + n\). | [
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6... | null | null |
aops_1180728 | Consier this equation as quadratic equation relatively variable $x$:
$x^2-x(y+2yz)+y^3z^2+y^3z=0$.
It's discriminant $D=y^2(2z+1)^2-4y^3z(z+1)$ should be a perfect square. As $D$ is divisible by $y^2$, then $D/y^2=(2z+1)^2-4yz(z+1)$ also is a perfect square, hence is at least 0.
But if $y>1$ then $4yz(z+1)>4z(z+1)+1=(... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $x,y,z$ are positive integers.Solve this equation\n$y^{3}z^{2}+y^{3}z+x^{2}-xy-2xyz=0$\n",
"content_html": "Let <img src=\"//latex.artofproblemsolving.com/a/c/c/accc80fdf164cef264f56a82b6f9f6add3... | Let \(x,y,z\) be positive integers. Solve the equation
\[
y^{3}z^{2}+y^{3}z+x^{2}-xy-2xyz=0.
\] | [
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numina_10090784 | We may limit the search to nonnegative integers $n$, since $n^{2}+15=(-n)^{2}+15$. Suppose there is a nonnegative integer $m$ such that
$$
\begin{aligned}
n^{2}+15 & =m^{2} \\
15 & =m^{2}-n^{2} \\
& =(m+n)(m-n) .
\end{aligned}
$$
Note that the factors $m+n$ and $m-n$ are not both negative, since their sum $2 m$ is no... | -7,-1,1,7 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Determine all integers $n$ for which $n^{2}+15$ is the square of an integer.
Remark. Because the problem asks you to "determine all integers $n$ ", you must verify that all the $n$ you find have the desired property, and moreover prove that these are the only such integers $n$. | [
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aops_1876044 | [b]Alternative solution:[/b]
We solve the following quadratic equation: $n^2+n+34-x^2=0$, so the delta must be perfect square, so we must have ${{y}^{2}}-4{{x}^{2}}=135\Leftrightarrow (y-2x)(y+2x)={{3}^{3}}5$, and now casework. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "What is the sum of all positive integers n such that n^2 + n + 34 is a perfect square?",
"content_html": "What is the sum of all positive integers n such that n^2 + n + 34 is a perfect square?",
... | What is the sum of all positive integers \(n\) such that \(n^2 + n + 34\) is a perfect square? | [
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numina_10038471 | 12.6. It is clear that $(x-1)(y-1)=x y-x-y+1=1$, therefore $x-1=y-1=1$, i.e., $x=y=2$. | 2 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 12.6. Solve the equation $x+y=x y$ in natural numbers. 12.7. Solve the equation $2 x y+3 x+y=0$ in integers. | [
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aops_1178595 | According to the given information there is an integer $k \neq 0$ s.t.
$(1) \;\; \frac{a^2}{2ab^2 - b^3 + 1} = k$,
yielding
$a^2 - 2kb^2a + k(b^3 - 1) = 0$,
which solution is
$(2) \;\; a = \frac{2kb^2 \pm d}{2}$
where $d \geq 0$ and
$(3) \;\; d^2 = (2kb^2)^2 - 4k(b^3 - 1)$.
Let us consider the following two ca... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $a, b >1$ naturals numbers,relatively prime. Prove that if $\\frac{a^2}{2ab^2 -b^3 +1}$ is integer number then $a=7 ,b=2$",
"content_html": "Let <img src=\"//latex.artofproblemsolving.com/f/3/e... | Let \(a,b>1\) be relatively prime natural numbers. Prove that if
\[
\frac{a^2}{2ab^2-b^3+1}
\]
is an integer, then \(a=7\) and \(b=2\). | [
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aops_2881732 | Solution from [i]Twitch Solves ISL[/i]:
The answer is $(-1,-2)$, $(-2,-1)$ and $(0,0)$. They work; we prove that's all.
Let $a = x+y$ and $b = xy$. Then this becomes \[ b(b^2-12b-12a+2) = (2a)^2 \implies 4 \cdot a^2 + 12b \cdot a - b(b^2-12b+2) = 0. \] As a quadratic in $a$, the discriminant is \[ (12b)^2 + 4 \cdot 4... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find all ordered pairs of integers $x,y$ such that $$xy(x^2y^2 - 12xy- 12x- 12y+2) = (2x + 2y)^2.$$\n\n[i]Proposed by Henry Jiang[/i]",
"content_html": "Find all ordered pairs of integers <img src=\"... | Find all ordered pairs of integers \(x,y\) such that
\[
xy\bigl(x^2y^2 - 12xy - 12x - 12y + 2\bigr) = (2x + 2y)^2.
\]
Proposed by Henry Jiang | [
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numina_10144144 | 1. Let $n^{2}+32 n+8=m^{3}(m>0)$. Then
$$
\begin{array}{l}
(n+m+16)(n-m+16) \\
=248=2^{3} \times 31 .
\end{array}
$$
Notice that, $n+m+16$ and $n-m+16$ have the same parity. Therefore, $n+m+16=124$ or 62. Accordingly, $n-m+16=2$ or 4.
Solving these, we get $n=47$ or 17. | 47or17 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 1. Find all positive integers $n$, such that $n^{2}+32 n+8$ is a perfect square. | [
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4... | null | null |
aops_1954604 | Quite easy...
[quote=Rizsgtp]Find all integers $x,y$ such that
$(x-y)^2+2y^2=27$.[/quote]
$\boxed{\text{My Solution :}}$
The expression is equivalent to $3y^2-2xy+(x^2-27)=0$
$\Delta=4x^2-12(x^2-27)=324-8x^2\ge0$
$\rightarrow{x\in{(0,1,-1,2,-2,...,6,-6)}}$
But also $324-8x^2\rightarrow{81-2x^2}$ must be a perfect s... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find all integers x,y such that \n(x-y)^2+2y^2=27.",
"content_html": "Find all integers x,y such that<br>\n(x-y)^2+2y^2=27.",
"post_id": 13500687,
"post_number": 1,
"post_time_unix"... | Find all integers \(x,y\) such that
\[
(x-y)^2 + 2y^2 = 27.
\] | [
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numina_10003673 | 27. How many integer solutions $(m ; n)$ does the equation param1 have?
| param1 | Answer |
| :---: | :---: |
| $m^{2}+9 m-27=n^{2}$ | 16 |
| $m^{2}+5 m-104=n^{2}$ | 18 |
| $m^{2}+3 m-279=n^{2}$ | 24 |
| $m^{2}+11 m-26=n^{2}$ | 18 |
| $m^{2}+7 m-139=n^{2}$ | 12 | | notfound | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 27. How many integer solutions ( $m ; n$ ) does the equation param1 have?
| param1 | Answer |
| :---: | :---: |
| $m^{2}+9 m-27=n^{2}$ | |
| $m^{2}+5 m-104=n^{2}$ | |
| $m^{2}+3 m-279=n^{2}$ | |
| $m^{2}+11 m-26=n^{2}$ | |
| $m^{2}+7 m-139=n^{2}$ | | | [
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numina_10000156 |
Solution. Substituting $s=x+y$ and $p=x y$ we get
$$
2 p^{2}-\left(s^{2}-3 s\right) p+19-s^{3}=0
$$
This is a quadratic equation in $p$ with discriminant $D=s^{4}+2 s^{3}+9 s^{2}-152$.
For each $s$ we have $D0$.
For $s \geqslant 11$ and $s \leqslant-8$ we have $D>\left(s^{2}+s+3\right)^{2}$ as this is equivalent t... | (2,1),(1,2),(-1,-20),(-20,-1) | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} |
N4. Find all integers $x, y$ such that
$$
x^{3}(y+1)+y^{3}(x+1)=19
$$
| [
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numina_10106310 | Solve: Regarding the original equation as a quadratic equation in $x$, it can be transformed into
$$
x^{2}+y x+\left(2 y^{2}-29\right)=0 \text {. }
$$
Since the equation has integer roots, the discriminant $\Delta$ must be a perfect square.
$$
\begin{array}{l}
\text { By } \Delta=y^{2}-4\left(2 y^{2}-29\right) \\
=-7 ... | 4 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 2 The equation about $x$ and $y$
$$
x^{2}+x y+2 y^{2}=29
$$
has ( ) groups of integer solutions $(x, y)$.
(A) 2
(B) 3
(C) 4
(D) infinitely many
(2009, National Junior High School Mathematics Competition) | [
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ours_25822 | To prove that there are no integers \( m \) and \( n \) satisfying the equation
\[
(m+n+2)^2 = 3(mn+1),
\]
we start by expanding both sides. The left-hand side expands to:
\[
(m+n+2)^2 = m^2 + 2mn + n^2 + 4m + 4n + 4.
\]
The right-hand side simplifies to:
\[
3(mn+1) = 3mn + 3.
\]
Equating both sides, we have:
\... | null | {
"competition": "serbian_mo",
"dataset": "Ours",
"posts": null,
"source": "bilten2015-1.md"
} | Prove that there are no integers \( m \) and \( n \) such that
\[
(m+n+2)^{2}=3(mn+1)
\] | [
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aops_225614 | [hide]$ 4a^2 + 4a + 1 - 29 = 4b^2$
$ (2a + 1)^2 - (2b)^2 = 29$
$ (2a + 2b + 1)(2a - 2b + 1) = 29$
$ \left \{ \begin{array}{c} 2a+2b+1=29 & 2a-2b+1=1 \end{array} \right \; \implies (a,b)= (7,7)$[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "find all $ a$ and $ b$ in $ N$ such \r\n$ a^2\\plus{}a\\minus{}7\\equal{}b^2$",
"content_html": "find all <img src=\"//latex.artofproblemsolving.com/2/5/5/255f65757f75ce300036173cb8e6f8f86dcfe90f.p... | Find all natural numbers \(a\) and \(b\) such that
\[
a^2 + a - 7 = b^2.
\] | [
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aops_1755086 | [hide=Solution]\begin{align*}
f(x)&=\frac x{x^2+6x+7}\\
(x^2+6x+7)f(x)&=x\\
f(x)x^2+(6f(x)-1)x+7f(x)&=0\\
x&=\frac{1-6f(x)\pm\sqrt{(1-6f(x))^2-28f(x)^2}}{2f(x)}\\
x&=\boxed{\frac{1-6f(x)\pm\sqrt{1-16f(x)+36f(x)^2}}{2f(x)}}
\end{align*} | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find x for f(x) = $\\frac{x}{x^2+6x+7} \\in Z$",
"content_html": "Find x for f(x) = <img src=\"//latex.artofproblemsolving.com/9/7/7/9777e7d131774ea24227aa52d04891531788ca91.png\" class=\"latex\" alt... | Find \(x\) such that \(f(x)=\dfrac{x}{x^2+6x+7}\in\mathbb{Z}\). | [
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aops_1830565 | [quote=huashiliao2020]I am not sure of a rigorous way to do this, and prove that this is the only solution. But perhaps note that xy increases faster than 3x+3y. So once we find one solution, increasing x and y too much will mean xy>3x+3y-1. Now, note that modulo 3, xy=2 mod 3, or one of the numbers is 2 mod 3, one of ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Product of two integers is $1$ less than three times of their sum. Find those integers.",
"content_html": "Product of two integers is <img src=\"//latex.artofproblemsolving.com/d/c/e/dce34f4dfb240614... | Find all integers \(x,y\) such that
\[
xy = 3(x+y)-1.
\] | [
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ours_26242 | Write the given equation in the form
$$
x^{2}(y+1)+x\left(2 y^{2}+3 y+3\right)+y^{3}+2 y^{2}+y+2=0.
$$
Note that for \(y=-1\), we have \(2x+2=0\), i.e., the pair \((-1,-1)\) is one solution. Now assume \(y \neq -1\). Then we can consider the equation as a quadratic equation in \(x\); let’s calculate its solutions:
$... | 41 | {
"competition": "serbian_mo",
"dataset": "Ours",
"posts": null,
"source": "bilten2018.md"
} | Determine how many equations
$$
y^{3}+x^{2} y+2 x y^{2}+x^{2}+3 x y+2 y^{2}+3 x+y+2=0
$$
have integer solutions \((x, y)\) for which \(|x| \leq 20\) and \(|y| \leq 18\). | [
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aops_3152536 | [hide=Partial Sol]
If we find a solution $(a,b,c)$ such that all $3$ variables are rational, we can just scale the solution such that all $3$ variables are integers.
Multiplying both sides by $abc$, we get $ca^2+b^2a+bc^2=73bca$.
$ca^2+(b^2-73bc)a+bc^2=0$
$D=b^4-146b^3c+5329b^2c^2-4bc^3=d^2$ for some integer $d$.
I thi... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find the positive integers $a,b,c$ such that \\[\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}=73.\\]",
"content_html": "Find the positive integers <img src=\"//latex.artofproblemsolving.com/a/5/b/a5b29b358e... | Find the positive integers \(a,b,c\) such that
\[
\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=73.
\] | [
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aops_362667 | Some particular solutions. The discriminant of $z^2 - (xy)z + (x^2 + y^2 -4) = 0$ is $\Delta = x^2y^2 - 4 (x^2 + y^2 -4) = (x^2-4)(y^2-4)$. Take $x = 2x'$ and $y=2y'$ to get $\Delta = 16(x'^2-1)(y'^2-1)$. Take $x' = A_1$ , $y' = A_2$ any particular solutions from the Pell equation $A^2 - dB^2 = 1$, to get $\Delta = 16d... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find all positive integer solutions to the equation $x^2+y^2+z^2-xyz=4$. The only progress I made was that there is an infinite number of solutions, and they are all reached from the solution $(x,x,2)$ for... | Find all positive integer solutions to the equation
\[
x^2+y^2+z^2-xyz=4.
\]
Is there any nice formula for all solutions of the equation? | [
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aops_600967 | Let us solve that in general, in real numbers. Amplify by $13$ and rearrange terms, to obtain the equivalent form $(13x-5)^2 + (13y-12)^2 = 0$. Thus the only real solution is $(x,y) = (5/13, 12/13)$. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Solve in integers $13(x^2+y^2+1)=10x+24y$.",
"content_html": "Solve in integers <span style=\"white-space:nowrap;\"><img src=\"//latex.artofproblemsolving.com/f/9/3/f93a9ea6e1c748435a728ddab7f16f910a... | Cleaned problem statement (LaTeX):
Solve in integers
\[
13(x^2+y^2+1)=10x+24y.
\] | [
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... | null | null |
aops_3434989 | [quote=Kempu33334]Try to factor it as a quadratic in $m$.[/quote]
This is actually a pretty common approach to problems like these. Specifically, rewrite the given equation as
\[
3m^2 + (4n + 2)m + n^2 = 0.\qquad(*)
\]
This is a quadratic with discriminant
\[
\Delta = (4n+2)^2 - 4\cdot 3n^2= 4(n^2 + 4n + 1).
\]
In or... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Integers $m$ and $n$ satisfy $3m^2 + 4mn + n^2 + 2m = 0$. What is the maximum possible value of $3m + n$?\n\n$\n(\\textbf{A})\\enspace {-2} \\qquad\n(\\textbf{B})\\enspace {-1} \\qquad\n(\\textbf{C})\\ensp... | Integers \(m\) and \(n\) satisfy
\[
3m^2 + 4mn + n^2 + 2m = 0.
\]
What is the maximum possible value of \(3m+n\)?
(A) \(-2\)\quad (B) \(-1\)\quad (C) \(0\)\quad (D) \(1\)\quad (E) \(2\) | [
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aops_494255 | [hide]Let $n^2+n+2009=k^2$, for some integer $k$. Then, the quadratic equation $n^2+n+2009-k^2=0$ has integer roots. Thus, its discriminant is a perfect square.
Therefore, $D=4k^2-8035=l^2$, for some ineger $l$. This implies that $(2k+l)(2k-1)=8035=5\cdot 1607$. Thus, $2k-l=1$ and $2k+l=8035$ or $2k-l=5$ and $2k+l=160... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find all $n$ positive integers such that $n^2+n+2009$ is a perfect square.",
"content_html": "Find all <img src=\"//latex.artofproblemsolving.com/1/7/4/174fadd07fd54c9afe288e96558c92e0c1da733a.png\" ... | Find all positive integers \(n\) such that \(n^2+n+2009\) is a perfect square. | [
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3... | null | null |
numina_10160460 | 30 Answer: (90)
$x^{2}+2(m+5) x+(100 m+9)=0 \Rightarrow x=-(m+5) \pm \sqrt{(m-45)^{2}-2009}$. This yields integer solutions if and only if $(m-45)^{2}-2009$ is a perfect square, say $n^{2}$.
Hence $(m-45)^{2}-n^{2}=2009=7^{2} \times 41 \Rightarrow$
$$
\begin{array}{c}
|m-45|+n=2009 \text { and }|m-45|-n=1 \text {, or }... | 90 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 30 Find the value of the smallest positive integer $m$ such that the equation
$$
x^{2}+2(m+5) x+(100 m+9)=0
$$
has only integer solutions. | [
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aops_1251266 | Since $S_n$ is a perfect square, let it be denoted by $m^2$.
$$m^2 = n^2+20n+12$$
$$m^2 =(n+10)^2-88$$
$$(n+10)^2 - m^2 = 88$$
$$(n+m+10)(n-m+10) = 88$$
88 has 8 factors,
n+m+10=88
n-m+10=1
n=34.5 (rejected)
n+m+10=44
n-m+10=2
n=13
n+m+10=22
n-m+10=4
n=3
n+m+10=11
n-m+10=8
n=-0.5 (rejected)
Hence, n=3 or 13 | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $S_n=n^{2} + 20n +12$, $n$ is a positive integer. What is the some of all possible values of $n$ for which $S_n$ is a perfect square? Please give full solution (pls).",
"content_html": "Let <span... | Let \(S_n = n^{2} + 20n + 12\), where \(n\) is a positive integer. Find the sum of all positive integers \(n\) for which \(S_n\) is a perfect square. Provide a full solution. | [
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aops_159963 | [hide]
$ (x+y)(x-y) = 81$
since x and y are integers, we know it will fall into one of the following factorizations
$ \pm1 * \pm81$
$ \pm3 * \pm27$
$ \pm9 * \pm9$
$ \pm27 * \pm3$
$ \pm81 * \pm1$
equating those back to the original x+y and x-y factors, we have (x,y) pairs of
$ (\pm41, \mp40)$
$ (\pm15, \mp12)$
$ (\p... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How many pairs of integers (x, y) are there for which x^2 − y^2 = 81?",
"content_html": "How many pairs of integers (x, y) are there for which x^2 − y^2 = 81?",
"post_id": 894514,
"post_n... | How many pairs of integers \((x,y)\) are there for which
\[
x^2 - y^2 = 81 ?
\] | [
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aops_206980 | Fix some $ x\in \mathbb{Z}$ and try to find $ y$. We have an second grade equality with variable $ y$: $ y^2 \plus{} y \minus{} (x^4 \plus{} x^3 \plus{} x^2 \plus{} x) \equal{} 0$.
To obtain $ y\in \mathbb{Z}$ we need that $ 1 \plus{} 4(x^4 \plus{} x^3 \plus{} x^2 \plus{} x)$ be an perfect square but note the followin... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find all solutions in integers of the equation:\r\n$ y^2\\plus{}y\\equal{}x^4\\plus{}x^3\\plus{}x^2\\plus{}x$ :ninja:",
"content_html": "Find all solutions in integers of the equation:<br>\n<img src=... | Find all integer solutions of the equation
\[
y^{2}+y = x^{4}+x^{3}+x^{2}+x.
\] | [
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... | null | null |
aops_515990 | Obviously, $b$ and $c$ are the roots of the equation $t^2-(4a+6)t+5(a^2+a+1)=0$. The discriminant, which is divisible by $4$, must be a perfect square:
$(4a+6)^2-20(a^2+a+1)=4k^2$ for some integer $k$
$-4a^2+28a+16=4k^2$
$(2k)^2+(2a-7)^2=65$
Hence a sum of two squares must be $65$. Checking among the numbers $1,4,9... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "let $a,b,c$ be positive integers such as: $5<a\\leq b\\leq c$, solve the system :\n\n$\\begin{cases}b+c=4a+6 \\\\ \\\\ bc =5(a^2+a+1) \\end{cases} $",
"content_html": "let <img src=\"//latex.artofpro... | Let \(a,b,c\) be positive integers such that \(5 < a \le b \le c\). Solve the system
\[
\begin{cases}
b+c=4a+6,\\[6pt]
bc=5(a^2+a+1).
\end{cases}
\] | [
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numina_10045280 | $\triangle$ Transform the equation by completing the square for the difference $y^{2}-2 y z$:
$$
5 x^{2}+\left(y^{2}-2 y z+z^{2}\right)+2 z^{2}=30, \quad 5 x^{2}+(y-z)^{2}+2 z^{2}=30
$$
From this, it is clear that the unknowns can only take values with small absolute values. Let's determine the constraints for $x$:
... | (1;5;0),(1;-5;0),(-1;5;0),(-1;-5;0) | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 812. Solve in integers $x, y$ and $z$ the equation
$$
5 x^{2} + y^{2} + 3 z^{2} - 2 y z = 30
$$ | [
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numina_10169649 | I. solution. Let's try to write the left side as a sum or difference of complete squares. Since $2 x y$ and $-2 x$ appear, let's see what remains if we write the expression $(x+y-1)^{2}$:
$$
\begin{aligned}
x^{2}-3 y^{2}+2 x y-2 x-10 y+20 & =(x+y-1)^{2}-4 y^{2}-8 y+19= \\
& =(x+y-1)^{2}-(2 y+2)^{2}+23
\end{aligned}
$$... | (19,-7),(-15,5),(7,5),(-3,-7) | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Solve the following equation over the set of integer pairs:
$$
x^{2}-3 y^{2}+2 x y-2 x-10 y+20=0
$$ | [
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52950,
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aops_1830256 | If $y=0$, then $x=0$, but we refuse trivial solution. So we can assume, that $y\neq 0$. Let's divide given equation by $y^2$ and let's put $t=x/y$ which is rational number. Then
$$t^2-nt+1=0.$$ If $\sqrt{\Delta}=\sqrt{n^2-4}$ is not rational, then the last equation hasn't solutions in rationals. But $n$ is an integer, ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "[hide][hide][hide]Find all n integers such that there exist non-zero-integers x,y ,so\n\nx^2-nxy+y^2=0 holds\n\nthere are 2 ways to solve it, post any one",
"content_html": "<span class=\"cmty-hide-... | Find all integers n for which there exist nonzero integers x,y satisfying
\[
x^2 - nxy + y^2 = 0.
\] | [
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23141,
40939,
... | null | null |
aops_492851 | [quote="maradona"]Find all naturel numbers $a$ and $b$ such that $a^4-49a^2+1=b^2$[/quote]
Let's substitute $a^2=t$, and write the equation in this form (as a quadratic equation):
$t^2-49t+1-b^2=0$
whose discriminant is:
$D=b^2-4ac=...=2397+4b^2$
Because we need natural solutions, the discriminant must be a perfe... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find all naturel numbers $a$ and $b$ such that $a^4-49a^2+1=b^2$",
"content_html": "Find all naturel numbers <img src=\"//latex.artofproblemsolving.com/c/7/d/c7d457e388298246adb06c587bccd419ea67f7e8.... | Find all natural numbers \(a\) and \(b\) such that
\[
a^4 - 49a^2 + 1 = b^2.
\] | [
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5030... | null | null |
aops_2048661 | [hide=Solution]If there was then we would have $(x-y)(x+y)=2003$. However $2003$ is prime, so our choices are narrowed down, but $(1002,1001)$ is a solution. Along with the negative values $(-1002,1001)$ or its opposite or $(-1002,-1001)$.
Full solutions are $(\pm 1002, \pm 1001)$.[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Does there exist a solution $x^2-y^2 = 2003$ in the integers ?",
"content_html": "Does there exist a solution <img src=\"//latex.artofproblemsolving.com/0/2/7/0270ae51358c594ade9c5abe0895f055ce035... | Does there exist a solution to
\[
x^2 - y^2 = 2003
\]
in integers \(x,y\)? | [
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4955... | null | null |
aops_1264121 | According to the question,
$$2a+1|a^2+4\implies a^2+4=k(2a+1)\implies a^2-2ak+4-k=0...........(1)$$
Now this is a quadratic of $a$,where $a$ is an integer.....∆$=c^2$
$$4k^2-4(4-k)=c^2\implies (2k+1)^2-c^2=17\implies (2k+1-c)(2k+1+c)=17$$
Now as $17$ is a prime the only solution comes as $$(2k+1=9,c=8) ,(2k+1=-9,c=... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find all integars a such that \n$\\frac{a^2 +4}{2a+1}$\nIs also an Integers..",
"content_html": "Find all integars a such that<br>\n<img src=\"//latex.artofproblemsolving.com/e/5/9/e59ccc309cb27bde4a... | Find all integers \(a\) such that
\[
\frac{a^2+4}{2a+1}
\]
is an integer. | [
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... | null | null |
aops_367370 | [quote="lssl"]Ver sorry it should be $ x^{2}-2xy+126y^{2}=2009 $ , I have edited .[/quote]
So $(x-y)^2+125y^2=2009$ and $y^2\le \frac {2009}{125}\sim 16.07$ and so $y^2\in\{0,1,4,9,16\}$
$y^2=0$ $\implies$ $(x-y)^2=2009$, impossible since $2009$ is not a perfect square
$y^2=1$ $\implies$ $(x-y)^2=1884$, impossible si... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find all pairs ($x,y$) of integers such that $x^2-2xy+126y^2=2009$.",
"content_html": "Find all pairs <span style=\"white-space:nowrap;\">(<img src=\"//latex.artofproblemsolving.com/b/6/4/b6400c6fe3f... | Find all pairs \((x,y)\) of integers such that
\[
x^2-2xy+126y^2=2009.
\] | [
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aops_490909 | [quote="xeroxia"]How many ordered pairs of positive integers $(x,y)$ are there such that $y^2-x^2=2y+7x+4$?
$ \textbf{(A)}\ 3
\qquad\textbf{(B)}\ 2
\qquad\textbf{(C)}\ 1
\qquad\textbf{(D)}\ 0
\qquad\textbf{(E)}\ \text{Infinitely many}
$[/quote]
[i]Solution[/i]: Write the equation : $x^2+7x+(4+2y-y^2)=0$
Thus, $\Delta... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How many ordered pairs of positive integers $(x,y)$ are there such that $y^2-x^2=2y+7x+4$?\n\n$ \\textbf{(A)}\\ 3\n\\qquad\\textbf{(B)}\\ 2\n\\qquad\\textbf{(C)}\\ 1\n\\qquad\\textbf{(D)}\\ 0\n\\qquad\\tex... | How many ordered pairs of positive integers \((x,y)\) are there such that
\[
y^2 - x^2 = 2y + 7x + 4?
\]
\[
\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 0 \qquad \textbf{(E)}\ \text{Infinitely many}
\] | [
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numina_10230443 | ### Part (a)
1. Given the equation:
\[
yx^2 + (y^2 - z^2)x + y(y - z)^2 = 0
\]
we can treat it as a quadratic equation in \(x\):
\[
yx^2 + (y^2 - z^2)x + y(y - z)^2 = 0
\]
The general form of a quadratic equation is \(ax^2 + bx + c = 0\), where:
\[
a = y, \quad b = y^2 - z^2, \quad c = y(y... | null | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "aops_forum"
} | Let $x,y,z$ be integer numbers satisfying the equality $yx^2+(y^2-z^2)x+y(y-z)^2=0$
a) Prove that number $xy$ is a perfect square.
b) Prove that there are infinitely many triples $(x,y,z)$ satisfying the equality.
I.Voronovich | [
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aops_103085 | If $y=1$, then $x+2+2z=xz$, $(x-2)(z-1)=4$, $(x,y,z)=(3,1,5),(4,1,3),(6,1,2)$
Consider $z=1$ the same way, $(x,y,z)=(3,5,1),(4,3,1),(6,2,1)$
Now, we consider the case $y>1, z>1$, then $yz\geq 4$
As $(y-1)(z-1)\geq 1$, we have $yz\geq y+z$
thus $x=\frac{2y+2z}{yz-1}\leq \frac{2yz}{yz-1}<3\to x=1,2$
If $x=... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find all triplets $(x,y,z) \\mathbb \\in{N}$ such that \r\n\r\n$x+2y+2z=xyz$",
"content_html": "Find all triplets <img src=\"//latex.artofproblemsolving.com/c/7/4/c74e25c0958584e5f2dcb37fe213477f1dc4... | Find all triplets \((x,y,z)\in\mathbb{N}^3\) such that
\[
x+2y+2z=xyz.
\] | [
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numina_10146433 | Solution 1 Let $(a, b)$ be a pair of positive integers satisfying the condition. Since
$$
k=\frac{a^{2}}{2 a b^{2}-b^{3}+1}>0,
$$
it follows that $2 a b^{2}-b^{3}+1>0, a>\frac{b}{2}-\frac{1}{2 b^{2}}$, thus $a \geqslant \frac{b}{2}$. Combining this with $k \geqslant 1$, we have
$$
a^{2} \geqslant b^{2}(2 a-b)+1,
$$
h... | (,b)=(2,1),(,2) | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 19 (IMO-44 Problem) Find all pairs of positive integers $(a, b)$ such that $\frac{a^{2}}{2 a b^{2}-b^{3}+1}$ is a positive integer. | [
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aops_480224 | WLOG $ x=max(x,y,z) $ .$x^2(1+y-z)+y^2(1+z-x)+z^2(1+x-y)=xy+xz+yz <=> (x-y)(z-y)(z-x)+\frac{(x-y)^{2}+(z-y)^{2}+(z-x)^{2}}{2}=0 $ If $y\ge z$ so $(x-y)(z-y)(z-x)+\frac{(x-y)^{2}+(z-y)^{2}+(z-x)^{2}}{2}\ge 0 $ Equality is attained $<=> x=y=z=n$ , $n$- is a integer.
If $x>z>y$. Let $ a=x-y , b=z-x ,a+b=z-y $ so $a>b$ a... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "solve this equation:(x,y and z are integer numbers) \nx^2(1+y-z)+y^2(1+z-x)+z^2(1+x-y)=xy+xz+yz",
"content_html": "solve this equation:(x,y and z are integer numbers)<br>\nx^2(1+y-z)+y^2(1+z-x)+z^2(... | Solve this equation (x, y, z are integers):
\[
x^2(1+y-z)+y^2(1+z-x)+z^2(1+x-y)=xy+xz+yz.
\] | [
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numina_10113385 | $$
\text { Three, let } \sqrt{\frac{a-x}{1+x}}=\frac{a-x^{2}}{1+x^{2}}=t \text {. }
$$
Then $t^{2} x+x+t^{2}-a=0$,
$$
(t+1) x^{2}+t-a=0 \text {. }
$$
$t \times(2)-x \times(1)$ gives
$t x^{2}-x^{2}-t^{2} x+t^{2}-a t+a x=0$
$\Rightarrow(t-x)(t+x-t x-a)=0$
$\Rightarrow t=x$ (discard) or $t=\frac{a-x}{1-x}$.
Thus $\frac{a... | 1 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Three. (25 points) If the pair of positive integers $(a, x)$ satisfies
$$
\sqrt{\frac{a-x}{1+x}}=\frac{a-x^{2}}{1+x^{2}} \neq x,
$$
find all positive integers $a$ that meet the requirement. | [
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numina_10153228 | $\begin{array}{l}n^{2}+5 n+13=n^{2}+5 n+2.5^{2}-2.5^{2}+13=(n+2.5)^{2}+6.75=m^{2} \text {, where } m \text { is an integer } \\ m^{2}-(n+2.5)^{2}=6.75 \Rightarrow(m+n+2.5)(m-n-2.5)=6.75 \Rightarrow(2 m+2 n+5)(2 n-2 m-5)=27 \\ \left\{\begin{array}{l}2 m+2 n+5=27 \\ 2 m-2 n-5=1\end{array} \text { or }\left\{\begin{array}... | 4 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | GS. 1 Given that $n$ is a positive integer. If $n^{2}+5 n+13$ is a perfect square, find the value of $n$. | [
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aops_2019563 | [quote=BobThePotato][hide=solution]\begin{align*}(x-y)^2=x+y &\implies x^2-(2y+1)x+(y^2-y)=0 \\
&\implies x=\frac{2y+1\pm\sqrt{(2y+1)^2-4(y^2-y)}}{2}=\frac{2y+1\pm\sqrt{8y+1}}{2}\end{align*}
Since $8y+1$ must be an odd perfect square, let it be equal to $(2z+1)^2=4z^2+4z+1,$ so that $y=\textstyle{\frac{1}{2}}(z^2+z).$ ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Looking for a solution to this question:\n\nFind all integer solutions to the equation\n$$\n(x - y)^2 = x + y\n$$",
"content_html": "Looking for a solution to this question:<br>\n<br>\nFind all integ... | Find all integer solutions to the equation
\[
(x - y)^2 = x + y.
\] | [
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aops_325858 | [hide] From the second equation, we have $ c(a\plus{}b)\equal{}23$ This means that either $ c\equal{}1$, abd $ a\plus{}b\equal{}23$, or $ c\equal{}23$, and $ a\plus{}b\equal{}1$ (since $ a$, $ b$ and $ c$ are positive integers0. However, the second scenario is invalid for that very reason, so $ c\equal{}1$ and $ a\plus... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Determine the number of triples $ (a,b,c)$ of positive integers which satisfy the simultaneous equations\r\n\\[ ab \\plus{} bc \\equal{} 44\\]\r\n\\[ ac \\plus{} bc \\equal{} 23\\]\r\n$ \\textbf{(A) } 0 \\... | Determine the number of triples \((a,b,c)\) of positive integers which satisfy the simultaneous equations
\[
ab+bc=44
\]
\[
ac+bc=23
\]
\[
\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4
\] | [
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aops_521947 | I think you mean POSITIVE integers: $c = 0$ kills the joy of this problem.
[hide="Long solution"]Important note: I use $a, b, c$ as mentioned in the problem, and $A, B, C$ as "quadratic equation coefficients" - they are NOT the same thing!
If we assume that all varaibles are integers, then plugging in $33 - bc = a$... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Given\n\na+bc=33\nb+ac=23\n\nSolve for c+ba.\n\n\nBTW, all the variables are intergers.",
"content_html": "Given<br>\n<br>\na+bc=33<br>\nb+ac=23<br>\n<br>\nSolve for c+ba.<br>\n<br>\n<br>\nBTW, all t... | Given
\[
a + bc = 33,
\qquad
b + ac = 23,
\]
where \(a,b,c\) are integers. Solve for \(c + ba\). | [
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aops_182062 | From $ a \leq b \leq c \leq 2a$ we have that $ a, b, c$ are sides of a triangle. I think that you can use that then exist $ x, y, z \geq 0$ such that $ a\equal{}y\plus{}z,$ $ b\equal{}z\plus{}x,$ and $ c\equal{}x\plus{}y$. :) | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$ 0<a\\le b\\le c\\le 2a$ with $ a\\plus{}b\\plus{}c\\equal{}4s$, show that $ abc\\ge 2s^3$.",
"content_html": "<img src=\"//latex.artofproblemsolving.com/2/8/0/280e6abfb5969cb306911a1ef89f60f3dd63d4... | Let \(0<a\le b\le c\le 2a\) with \(a+b+c=4s\). Show that \(abc\ge 2s^3\). | [
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aops_2868150 | $ P(1)<0,P(2)=1>0,P(3)=-4<0,P(4)=1>0$ so we have $3$ roots $1<a<2<b<3<c<4$
a) $a+b+c=8$ and $c<4$ so $a+b>4>c$ so $a,b,c$ form triangle.
b) $S^2=\frac{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}{4}=\frac{(a+b+c)(-(a+b+c)^3+4(a+b+c)(ab+bc+ca)-8abc}{4}=\frac{8(-8^3+4*8*20-8\frac{79}{5})}{4}=\frac{16}{5} \to S=\frac{4}{\sqrt{5}}$
c) ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Given the polynomial $P(x) = 5x^3 - 40x^2 + 100x - 79$ which has 3 roots $a,b,c$ \na) Prove that $a,b,c$ form an triangle \nb) Find the area of the triangle from a) \nc) Prove that the length of the 3 medi... | Given the polynomial \(P(x)=5x^{3}-40x^{2}+100x-79\) which has three roots \(a,b,c\).
a) Prove that \(a,b,c\) form a triangle.
b) Find the area of the triangle from (a).
c) Prove that the lengths of the three medians of that triangle form an obtuse triangle. | [
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numina_10114810 | From the Law of Sines, we get $\frac{b}{\sin B}=\frac{c}{\sin C} \Rightarrow \sin B=2 \sin C$, thus $\cos B+\frac{1}{2} \sin B=1$. Substituting $\sin ^{2} B+\cos ^{2} B=1$ yields $\sin B=\frac{4}{5}, \cos B=\frac{3}{5}, \sin C=\frac{2}{5}, \cos C=\frac{\sqrt{21}}{5}$.
Therefore, $\cos A=-\cos (B+C)=\frac{4}{5} \cdot \f... | \frac{2\sqrt{21}+3}{5} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 9. Given $\triangle A B C$ satisfies $A B=1, A C=2, \cos B+\sin C=1$, find the length of side $B C$.
---
The translation is provided as requested, maintaining the original format and line breaks. | [
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aops_442079 | It is not mentioned in what dimension the rays live, and it matters. Say they lie in the plane. The largest size of a set of rays in $\mathbb{R}^2$ making pairwise obtuse angles is easily seen to be $3$. Therefore if we consider the graph having the $n$ rays as vertices, with edges connecting the rays making an obtuse ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Consider $n \\ge 3$ rays that originate from a point. What is the maximum number of obtuse angles they can form? (The angle between any two rays is taken to be less than or equal to $180^\\circ$.)\n\n(Defi... | Consider \(n\ge 3\) rays that originate from a point. What is the maximum number of obtuse angles they can form? (The angle between any two rays is taken to be less than or equal to \(180^\circ\).)
Definition. An obtuse angle is an angle larger than a right angle and smaller than a straight angle (between \(90^\circ\)... | [
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aops_1653098 | [quote=Mathkeys]
Please explain with and without the use of combinations. Thanks![/quote]
if you draw a diagram, then you get 10 points on a circle, no three are collinear, so you can connect point AB,AC,AD..., so 9 for the first 1, the 8 for the second cause you already counted AB and counting BA would be overcounting... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Hi everyone, \n\nI need a clear answer on this one please.\n\nIf there are 10 points on a circle, how many line segments can be made by connecting any two points?\n\nPlease explain with and without the use... | If there are 10 points on a circle, how many line segments can be made by connecting any two points?
Please explain with and without the use of combinations. | [
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numina_10087624 | ## Solution.
Notice that the $n$-th term of the given sum is of the form $a_{n}=\frac{2 n+1}{n^{2} \cdot(n+1)^{2}}$.
Furthermore, the term $a_{n}$ can be written in the following form:
$$
\begin{aligned}
& a_{n}=\frac{2 n+1}{n^{2} \cdot(n+1)^{2}}=\frac{n}{n^{2} \cdot(n+1)^{2}}+\frac{n+1}{n^{2} \cdot(n+1)^{2}}=\frac{... | \frac{9999}{10000} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | ## Task B-4.5.
Calculate the sum $\frac{3}{1^{2} \cdot 2^{2}}+\frac{5}{2^{2} \cdot 3^{2}}+\frac{7}{3^{2} \cdot 4^{2}}+\cdots+\frac{199}{99^{2} \cdot 100^{2}}$. | [
36846,
2184,
18145,
27669,
7290,
39069,
54572,
60673,
9553,
35912,
30187,
47610,
23998,
32608,
26410,
34365,
62211,
57937,
64330,
19047,
51474,
41360,
7490,
70523,
56167,
65728,
49878,
30571,
26267,
58118,
44963,
39875,
46259,
30081,
52756,
44680... | null | null |
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