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Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 13"
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==Solution==
 
==Solution==
 
Assume temporarily that <math>p \neq 2</math>. Then, <math>p</math> and <math>n</math> are both odd which implies that the numerator is odd and the denominator is even. Since an even number cannot divide an odd number, we have a contradiction. Since our claim is incorrect, <math>p=2</math> and we now wish to make <cmath>\frac{2^{n+2023}}{(n+2)^2}</cmath> an integer. We see that the denominator is always odd, while the numerator is always even. Thus, the only possible values for <math>n</math> are when the denominator is <math>1,-1</math>, which implies <math>n=-1,-3</math>. These correspond with <math>p=2</math>, so <math>pn=-2,-6</math> for an answer of <math>-8</math>. ~samrocksnature
 
Assume temporarily that <math>p \neq 2</math>. Then, <math>p</math> and <math>n</math> are both odd which implies that the numerator is odd and the denominator is even. Since an even number cannot divide an odd number, we have a contradiction. Since our claim is incorrect, <math>p=2</math> and we now wish to make <cmath>\frac{2^{n+2023}}{(n+2)^2}</cmath> an integer. We see that the denominator is always odd, while the numerator is always even. Thus, the only possible values for <math>n</math> are when the denominator is <math>1,-1</math>, which implies <math>n=-1,-3</math>. These correspond with <math>p=2</math>, so <math>pn=-2,-6</math> for an answer of <math>-8</math>. ~samrocksnature
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==Solution 2==
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Suppose <math>p</math> is odd. We have that our expression can never be an integer, so <math>p=2</math>. Now, <math>\frac{2^{n+2023}}{(2+n)^2}</math> is an integer, implying <math>(2+n)</math> is a perfect power of <math>2</math>. Now, we have <math>n</math> is even if <math>(2+n) \ge 1</math>, and <math>n=\{-1,-3 \}</math> are the only values that work when testing for odd <math>n</math>. The answer is <math>2(-1-3)=\boxed{-8}</math>
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~Geometry285
  
 
==See also==
 
==See also==
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#[[2021 JMPSC Invitational Problems|Other 2021 JMPSC Invitational Problems]]
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#[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]]
−
#[[2021 JMPSC Invitational Answer Key|2021 JMPSC Invitational Answer Key]]
+
#[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]]
 
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 
{{JMPSC Notice}}
 
{{JMPSC Notice}}

Latest revision as of 20:17, 11 July 2021

Problem

Let $p$ be a prime and $n$ be an odd integer (not necessarily positive) such that \[\dfrac{p^{n+p+2021}}{(p+n)^2}\] is an integer. Find the sum of all distinct possible values of $p \cdot n$.

Solution

Assume temporarily that $p \neq 2$. Then, $p$ and $n$ are both odd which implies that the numerator is odd and the denominator is even. Since an even number cannot divide an odd number, we have a contradiction. Since our claim is incorrect, $p=2$ and we now wish to make \[\frac{2^{n+2023}}{(n+2)^2}\] an integer. We see that the denominator is always odd, while the numerator is always even. Thus, the only possible values for $n$ are when the denominator is $1,-1$, which implies $n=-1,-3$. These correspond with $p=2$, so $pn=-2,-6$ for an answer of $-8$. ~samrocksnature

Solution 2

Suppose $p$ is odd. We have that our expression can never be an integer, so $p=2$. Now, $\frac{2^{n+2023}}{(2+n)^2}$ is an integer, implying $(2+n)$ is a perfect power of $2$. Now, we have $n$ is even if $(2+n) \ge 1$, and $n=\{-1,-3 \}$ are the only values that work when testing for odd $n$. The answer is $2(-1-3)=\boxed{-8}$

~Geometry285

See also

  1. Other 2021 JMPSC Invitationals Problems
  2. 2021 JMPSC Invitationals Answer Key
  3. All JMPSC Problems and Solutions

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