Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 13"

(Created page with "==Problem== Let <math>p</math> be a prime and <math>n</math> be an odd integer (not necessarily positive) such that <cmath>\dfrac{p^{n+p+2021}}{(p+n)^2}</cmath> is an integer....")
 
m
 
(2 intermediate revisions by 2 users not shown)
Line 4: Line 4:
 
==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
 +
 +
==Solution 2==
 +
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>
 +
 +
~Geometry285
 +
 +
==See also==
 +
#[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]]
 +
#[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]]
 +
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 +
{{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

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Invitationals_Problems/Problem_13&oldid=158151"