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Difference between revisions of "2017 AIME I Problems/Problem 9"

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Therefore, <math>a_n</math> is divisible by 99 if and only if <math>\frac{1}{2}(n+10)(n-9)</math> is divisible by 99, so <math>(n+10)(n-9)</math> needs to be divisible by 9 and 11. Assume that <math>n+10</math> is a multiple of 11. Writing out a few terms, <math>n=12, 23, 34, 45</math>, we see that <math>n=45</math> is the smallest <math>n</math> that works in this case. Next, assume that <math>n-9</math> is a multiple of 11. Writing out a few terms, <math>n=20, 31, 42, 53</math>, we see that <math>n=53</math> is the smallest <math>n</math> that works in this case. The smallest <math>n</math> is <math>\boxed{45}</math>.
 
Therefore, <math>a_n</math> is divisible by 99 if and only if <math>\frac{1}{2}(n+10)(n-9)</math> is divisible by 99, so <math>(n+10)(n-9)</math> needs to be divisible by 9 and 11. Assume that <math>n+10</math> is a multiple of 11. Writing out a few terms, <math>n=12, 23, 34, 45</math>, we see that <math>n=45</math> is the smallest <math>n</math> that works in this case. Next, assume that <math>n-9</math> is a multiple of 11. Writing out a few terms, <math>n=20, 31, 42, 53</math>, we see that <math>n=53</math> is the smallest <math>n</math> that works in this case. The smallest <math>n</math> is <math>\boxed{45}</math>.
  
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==See also==
 
{{AIME box|year=2017|n=I|num-b=8|num-a=10}}
 
{{AIME box|year=2017|n=I|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:04, 8 March 2017

Problem 9

Let $a_{10} = 10$, and for each integer $n >10$ let $a_n = 100a_{n - 1} + n$. Find the least $n > 10$ such that $a_n$ is a multiple of $99$.

Solution

Writing out the recursive statement for $a_n, a_{n-1}, \dots, a_{10}$ and summing them gives \[a_n+\dots+a_{10}=100(a_{n-1}+\dots+a_{10})+n+\dots+10\] Which simplifies to \[a_n=99(a_{n-1}+\dots+a_{10})+\frac{1}{2}(n+10)(n-9)\] Therefore, $a_n$ is divisible by 99 if and only if $\frac{1}{2}(n+10)(n-9)$ is divisible by 99, so $(n+10)(n-9)$ needs to be divisible by 9 and 11. Assume that $n+10$ is a multiple of 11. Writing out a few terms, $n=12, 23, 34, 45$, we see that $n=45$ is the smallest $n$ that works in this case. Next, assume that $n-9$ is a multiple of 11. Writing out a few terms, $n=20, 31, 42, 53$, we see that $n=53$ is the smallest $n$ that works in this case. The smallest $n$ is $\boxed{45}$.

See also

2017 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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