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

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

Revision as of 16:21, 12 March 2020

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Problem

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 $\boxed{407}$.

See Also

2020 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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