Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AIME I Problems/Problem 10"
 
(9 intermediate revisions by 7 users not shown)
Line 1: Line 1:
−
 
 
== Problem ==
 
== Problem ==
 
Let <math>m</math> and <math>n</math> be positive integers satisfying the conditions
 
Let <math>m</math> and <math>n</math> be positive integers satisfying the conditions
Line 32: Line 31:
 
<math>275+121</math> is divisible by 2, 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>.
 
<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>.
 +
 +
==Solution 3 (Official MAA)==
 +
Let <math>m</math> and <math>n</math> be positive integers where <math>m^m</math> is a multiple of <math>n^n</math> and <math>m</math> is not a multiple of <math>n</math>. If a prime <math>p</math> divides <math>n</math>, then <math>p</math> divides <math>n^n</math>, so it also divides <math>m^m</math>, and thus <math>p</math> divides <math>m</math>. Therefore any prime <math>p</math> dividing <math>n</math> also divides both <math>m</math> and <math>k = m + n</math>. Because <math>k</math> is relatively prime to <math>210=2\cdot3\cdot5\cdot7</math>, the primes <math>2</math>, <math>3</math>, <math>5</math>, and <math>7</math> cannot divide <math>n</math>. Furthermore, because <math>m</math> is divisible by every prime factor of <math>n</math>, but <math>m</math> is not a multiple of <math>n</math>, the integer <math>n</math> must be divisible by the square of some prime, and that prime must be at least <math>11</math>. Thus <math>n</math> must be at least <math>11^2 = 121</math>.
 +
 +
If <math>n=11^2</math>, then <math>m</math> must be a multiple of <math>11</math> but not a multiple of <math>121</math>, and <math>m^m</math> must be divisible by <math>n^n = 121^{121} = 11^{242}</math>. Therefore <math>m</math> must be a multiple of <math>11</math> that is greater than <math>242</math>. Let <math>m = 11m_0</math>, with <math>m_0 > 22</math>. Then <math>k = m + n = 11(m_0 + 11)</math>. The least <math>m_0 > 22</math> for which <math>m_0 + 11</math> is not divisible by any of the primes <math>2</math>, <math>3</math>, <math>5</math>, or <math>7</math> is <math>m_0 = 26</math>, giving the prime <math>m_0 + 11 = 37</math>. Hence the least possible <math>k</math> when <math>n = 121</math> is <math>k = 11 \cdot 37 = 407</math>.
 +
 +
It remains to consider other possible values for <math>n</math>. If <math>n = 13^2 = 169</math>, then <math>m</math> must be divisible by <math>13</math> but not <math>169</math>, and <math>m^m</math> must be a multiple of <math>n^n = 169^{169} = 13^{338}</math>, so <math>m > 338</math>. Then <math>k = m + n > 169 + 338 = 507</math>. All other possible values for <math>n</math> have <math>n \ge 242</math>, and in this case <math>m > n \ge 242</math>, so <math>k \ge 2 \cdot 242 = 484</math>. Hence no greater values of <math>n</math> can produce lesser values for <math>k</math>, and the least possible <math>k</math> is indeed <math>407</math>.
  
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/Z47NRwNB-D0
 
https://youtu.be/Z47NRwNB-D0
 +
 +
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=FQSiQChGjpI&list=PLLCzevlMcsWN9y8YI4KNPZlhdsjPTlRrb&index=7 ~ MathEx
  
 
==See Also==
 
==See Also==

Latest revision as of 19:00, 31 October 2021

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

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 3 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_10&oldid=164414"