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"

m (→‎Solution 0)
m (→‎Solution 1)
Line 14: Line 14:
 
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
 
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 1 ==
+
== 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.
 
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.
  

Revision as of 16:50, 13 March 2020

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 0

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 $\boxed{407}$.

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=119342"