Difference between revisions of "2020 AIME I Problems/Problem 10"
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== Solution == | == Solution == | ||
+ | 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 we are inspired to take to be , the lowest prime not dividng , or . Now, there are factors of , so , and then for . Now, . Noting is the minimal that satisfies this, we get . Thus, it is easy to verify this is minimal and we get .
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
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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