Difference between revisions of "2020 AIME I Problems/Problem 10"
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− | + | == 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 1== | ||
+ | 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 dividing <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>. ~awang11 | ||
+ | |||
+ | == Solution 2 == | ||
+ | Assume for the sake of contradiction that <math>n</math> is a multiple of a single digit prime number, then <math>m</math> must also be a multiple of that single digit prime number to accommodate for <math>n^n | m^m</math>. However that means that <math>m+n</math> is divisible by that single digit prime number, which violates <math>\gcd(m+n,210) = 1</math>, so contradiction. | ||
+ | |||
+ | <math>n</math> is also not 1 because then <math>m</math> would be a multiple of it. | ||
+ | |||
+ | Thus, <math>n</math> is a multiple of 11 and/or 13 and/or 17 and/or... | ||
+ | |||
+ | Assume for the sake of contradiction that <math>n</math> has at most 1 power of 11, at most 1 power of 13...and so on... | ||
+ | Then, for <math>n^n | m^m</math> to be satisfied, <math>m</math> must contain at least the same prime factors that <math>n</math> has. This tells us that for the primes where <math>n</math> has one power of, <math>m</math> also has at least one power, and since this holds true for all the primes of <math>n</math>, <math>n|m</math>. Contradiction. | ||
+ | |||
+ | Thus <math>n</math> needs more than one power of some prime. | ||
+ | The obvious smallest possible value of <math>n</math> now is <math>11^2 =121</math>. | ||
+ | Since <math>121^{121}=11^{242}</math>, we need <math>m</math> to be a multiple of 11 at least <math>242</math> that is not divisible by <math>121</math> and most importantly, <math>\gcd(m+n,210) = 1</math>. | ||
+ | <math>242</math> is divisible by <math>121</math>, out. | ||
+ | <math>253+121</math> is divisible by 2, out. | ||
+ | <math>264+121</math> is divisible by 5, out. | ||
+ | <math>275+121</math> is divisible by 2, out. | ||
+ | <math>286+121=37\cdot 11</math> and satisfies all the conditions in the given problem, and the next case <math>n=169</math> will give us at least <math>169\cdot 3</math>, so we get <math>\boxed{407}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/Z47NRwNB-D0 | ||
− | |||
− | == Solution == | + | ==Video Solution== |
+ | https://www.youtube.com/watch?v=FQSiQChGjpI&list=PLLCzevlMcsWN9y8YI4KNPZlhdsjPTlRrb&index=7 ~ MathEx | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2020|n=I|num-b=9|num-a=11}} | {{AIME box|year=2020|n=I|num-b=9|num-a=11}} | ||
+ | [[Category:Intermediate Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:18, 25 November 2020
Problem
Let and be positive integers satisfying the conditions
is a multiple of and
is not a multiple of
Find the least possible value of
Solution 1
Taking inspiration from we are inspired to take to be , the lowest prime not dividing , 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 . ~awang11
Solution 2
Assume for the sake of contradiction that is a multiple of a single digit prime number, then must also be a multiple of that single digit prime number to accommodate for . However that means that is divisible by that single digit prime number, which violates , so contradiction.
is also not 1 because then would be a multiple of it.
Thus, is a multiple of 11 and/or 13 and/or 17 and/or...
Assume for the sake of contradiction that has at most 1 power of 11, at most 1 power of 13...and so on... Then, for to be satisfied, must contain at least the same prime factors that has. This tells us that for the primes where has one power of, also has at least one power, and since this holds true for all the primes of , . Contradiction.
Thus needs more than one power of some prime. The obvious smallest possible value of now is . Since , we need to be a multiple of 11 at least that is not divisible by and most importantly, . is divisible by , out. is divisible by 2, out. is divisible by 5, out. is divisible by 2, out. and satisfies all the conditions in the given problem, and the next case will give us at least , so we get .
Video Solution
Video Solution
https://www.youtube.com/watch?v=FQSiQChGjpI&list=PLLCzevlMcsWN9y8YI4KNPZlhdsjPTlRrb&index=7 ~ MathEx
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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