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== | ||
+ | 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== | ==See also== | ||
− | #[[2021 JMPSC | + | #[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]] |
− | #[[2021 JMPSC | + | #[[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 21:17, 11 July 2021
Contents
Problem
Let be a prime and
be an odd integer (not necessarily positive) such that
is an integer. Find the sum of all distinct possible values of
.
Solution
Assume temporarily that . Then,
and
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,
and we now wish to make
an integer. We see that the denominator is always odd, while the numerator is always even. Thus, the only possible values for
are when the denominator is
, which implies
. These correspond with
, so
for an answer of
. ~samrocksnature
Solution 2
Suppose is odd. We have that our expression can never be an integer, so
. Now,
is an integer, implying
is a perfect power of
. Now, we have
is even if
, and
are the only values that work when testing for odd
. The answer is
~Geometry285
See also
- Other 2021 JMPSC Invitationals Problems
- 2021 JMPSC Invitationals Answer Key
- All JMPSC Problems and Solutions
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.