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

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We essentially have that <math>\gcd(m,n) \notin \{1, m, n\},</math> or any factor of <math>210.</math>
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We essentially have that <math>\gcd(m,n) \notin \{1, m, n\},</math> or the set of factors of <math>210.</math>
  
  

Revision as of 18:35, 16 August 2025

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 1

Taking inspiration from $4^4 \mid 10^{10}$ we are inspired to take $n$ to be $p^2$, the lowest prime not dividing $210$, 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

Solution 2 (step by step)

We essentially have that $\gcd(m,n) \notin \{1, m, n\},$ or the set of factors of $210.$


This implies that one number must be odd and the other must be even so that they don’t add up to a multiple of $2.$ If $n$ is even and $m$ is odd it would be impossible for $m^m$ to be a multiple of $n^n,$ so $m$ must be even and $n$ must be odd.

Now we need to choose the prime factor for $m$ and $n$ to share. It can't be $2,3,5,$ or $7$ so the smallest option is $11.$

So far we have $m = 22$ and $n = 11.$ We need to add more factors so that $m$ is not a multiple of $n$ but $m^m$ is a multiple of $n^n.$ Note that no matter how many additional factors we add to $m,$ it will always be a multiple of $11,$ so we have to add another factor to $n$ to make sure it doesn't divide $m.$ Again the smallest option is $11,$ so $n$ becomes $121 \implies n^n = 11^{242},$ and $m^m = 2^{22} \cdot 11^{22}.$ All we need to do now is significantly increase the value of $m$ so that the exponent on $11$ becomes larger than $242.$ If we added another factor of $11$ to $m$ then it would be a multiple of $n,$ so the next smallest option is $13.$ Then $m = 22\cdot 13 = 286.$ $m^m$ becomes $2^{286} \cdot 11^{286} \cdot 13^{286}$ which satisfies the problem's condition.

Therefore, the least possible value of $m + n$ is $121 + 286 = \boxed{407}.$

~grogg007

Solution 3

Assume for the sake of contradiction that $n$ is a multiple of a single digit prime number, then $m$ must also be a multiple of that single digit prime number to accommodate for $n^n | m^m$. However that means that $m+n$ is divisible by that single digit prime number, which violates $\gcd(m+n,210) = 1$, so contradiction.

$n$ is also not 1 because then $m$ would be a multiple of it.

Thus, $n$ is a multiple of 11 and/or 13 and/or 17 and/or...

Assume for the sake of contradiction that $n$ has at most 1 power of 11, at most 1 power of 13...and so on... Then, for $n^n | m^m$ to be satisfied, $m$ must contain at least the same prime factors that $n$ has. This tells us that for the primes where $n$ has one power of, $m$ also has at least one power, and since this holds true for all the primes of $n$, $n|m$. Contradiction.

Thus $n$ needs more than one power of some prime. The obvious smallest possible value of $n$ now is $11^2 =121$. Since $121^{121}=11^{242}$, we need $m$ to be a multiple of 11 at least $242$ that is not divisible by $121$ and most importantly, $\gcd(m+n,210) = 1$. $242$ is divisible by $121$, out. $253+121$ is divisible by 2, out. $264+121$ is divisible by 5, out. $275+121$ is divisible by 2, out. $286+121=37\cdot 11$ and satisfies all the conditions in the given problem, and the next case $n=169$ will give us at least $169\cdot 3$, so we get $\boxed{407}$.

Solution 4 (Official MAA)

Let $m$ and $n$ be positive integers where $m^m$ is a multiple of $n^n$ and $m$ is not a multiple of $n$. If a prime $p$ divides $n$, then $p$ divides $n^n$, so it also divides $m^m$, and thus $p$ divides $m$. Therefore any prime $p$ dividing $n$ also divides both $m$ and $k = m + n$. Because $k$ is relatively prime to $210=2\cdot3\cdot5\cdot7$, the primes $2$, $3$, $5$, and $7$ cannot divide $n$. Furthermore, because $m$ is divisible by every prime factor of $n$, but $m$ is not a multiple of $n$, the integer $n$ must be divisible by the square of some prime, and that prime must be at least $11$. Thus $n$ must be at least $11^2 = 121$.

If $n=11^2$, then $m$ must be a multiple of $11$ but not a multiple of $121$, and $m^m$ must be divisible by $n^n = 121^{121} = 11^{242}$. Therefore $m$ must be a multiple of $11$ that is greater than $242$. Let $m = 11m_0$, with $m_0 > 22$. Then $k = m + n = 11(m_0 + 11)$. The least $m_0 > 22$ for which $m_0 + 11$ is not divisible by any of the primes $2$, $3$, $5$, or $7$ is $m_0 = 26$, giving the prime $m_0 + 11 = 37$. Hence the least possible $k$ when $n = 121$ is $k = 11 \cdot 37 = 407$.

It remains to consider other possible values for $n$. If $n = 13^2 = 169$, then $m$ must be divisible by $13$ but not $169$, and $m^m$ must be a multiple of $n^n = 169^{169} = 13^{338}$, so $m > 338$. Then $k = m + n > 169 + 338 = 507$. All other possible values for $n$ have $n \ge 242$, and in this case $m > n \ge 242$, so $k \ge 2 \cdot 242 = 484$. Hence no greater values of $n$ can produce lesser values for $k$, and the least possible $k$ is indeed $407$.

Video Solution

https://youtu.be/Z47NRwNB-D0


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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