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Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 15"

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Counting all <math>2</math>, <math>3</math>, and <math>4</math> digit combinations and then permuting only those up to <math>2045</math>, we find that there are 186 numbers whose sums are either <math>11</math> or <math>22</math>. We need not account for the sum 33, as it is not achievable with a <math>2</math> as the lowest digit. Since there are a total of <math>2048</math> numbers and <math>186</math> that work, we get <math>186/2048</math> or <math>93/1024</math>. Our sum is then <math>93 + 1024 = 1117</math>. The last three digits are <math>117</math>.
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Counting all <math>2</math>, <math>3</math>, and <math>4</math> digit combinations and then permuting only those up to <math>2045</math>, we find that there are 186 numbers whose sums are either <math>11</math> or <math>22</math>. We need not account for the sum 33, as it is not achievable with a <math>2</math> as the lowest digit. Since there are a total of <math>2048</math> numbers and <math>186</math> that work, we get <math>186/2048</math> or <math>93/1024</math>. Our sum is then <math>93 + 1024 = 1117</math>. The last three digits are <math>\boxed{117}</math>.
  
  

Revision as of 13:52, 25 April 2009

Problem

Let $S$ be the set of integers $0,1,2,...,10^{11}-1$. An element $x\in S$ (in) is chosen at random. Let $\star (x)$ denote the sum of the digits of $x$. The probability that $\star(x)$ is divisible by 11 is $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Compute the last 3 digits of $m+n$

Solution

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Counting all $2$, $3$, and $4$ digit combinations and then permuting only those up to $2045$, we find that there are 186 numbers whose sums are either $11$ or $22$. We need not account for the sum 33, as it is not achievable with a $2$ as the lowest digit. Since there are a total of $2048$ numbers and $186$ that work, we get $186/2048$ or $93/1024$. Our sum is then $93 + 1024 = 1117$. The last three digits are $\boxed{117}$.


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