Difference between revisions of "2020 AIME I Problems/Problem 10"

Revision as of 16:35, 12 March 2020

Note: Please do not post problems here until after the AIME.

Problem

Let $m$ and $n$ be positive integers satisfying the conditions

$\quad\bullet\ \gcd(m+n,210)=1,$

$\quad\bullet\ m^m$ is a multiple of $n^n,$ and

$\quad\bullet\ m$ is not a multiple of $n.$

Find the least possible value of $m+n.$

Solution

Taking inspiration from $4^4 \mid 10^{10}$ we are inspired to take $n$ to be $p^2$ , the lowest prime not dividng $210$ , or $11 \implies n = 121$ . Now, there are $242$ factors of $11$ , so $11^{242} \mid m^m$ , and then $m = 11k$ for $k \geq 22$ . Now, $\gcd(m+n, 210) = \gcd(11+k,210) = 1$ . Noting $k = 26$ is the minimal that satisfies this, we get $(n,m) = (121,286)$ . Thus, it is easy to verify this is minimal and we get $\boxed{407}$ . ~awang11

@@ Line 2: / Line 2: @@
 == Problem ==
+Let <math>m</math> and <math>n</math> be positive integers satisfying the conditions
+<math>\quad\bullet\ \gcd(m+n,210)=1,</math>
+<math>\quad\bullet\ m^m</math> is a multiple of <math>n^n,</math> and
+<math>\quad\bullet\ m</math> is not a multiple of <math>n.</math>
+Find the least possible value of <math>m+n.</math>
 == Solution ==

2020 AIME I (Problems • Answer Key • Resources)
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Difference between revisions of "2020 AIME I Problems/Problem 10"

Revision as of 16:35, 12 March 2020

Problem

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