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Difference between revisions of "2013 Mock AIME I Problems/Problem 2"

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== Solution ==
 
== Solution ==
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<math>\boxed{005}</math>.
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Because <math>a|b</math>, let <math>b=an</math>, where <math>n</math> is a positive integer. Because <math>c-a=10</math>, <math>c=a+10</math>, so <math>(b+1)|(a+10)</math> and thus <math>(an+1)|(a+10)</math>. Now, let <math>a+10=m(an+1)</math>, where <math>m</math> is another positive integer. Thus, <math>a=\tfrac{10-m}{mn-1}</math>. Because the ordered pair <math>(m,n)</math> uniquely determines values of <math>a</math>, <math>b</math>, and <math>c</math>, the desired number of triples <math>(a,b,c)</math> that fit the constaints of the problem equals the number of positive integer pairs <math>(m,n)</math> that force <math>a=\tfrac{10-m}{mn-1}</math> and, consequently, <math>b</math> and <math>c</math>, to be positive integers.
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Starting with <math>n=1</math>, by listing out fractions of the form <math>\tfrac{10-m}{mn-1}</math> and seeing if they simplify to positive integers, we see that the only possible values of <math>m</math> are <math>2</math> and <math>4</math>. Likewise, for <math>n=2</math>, <math>m</math> must be <math>1</math>. For <math>n=4</math>, <math>m=1</math>, and for <math>n=10</math>, <math>m=1</math>. No  other values of <math>n</math> yield positive integer values of <math>m</math>. Thus, because there are <math>5</math> ordered pairs <math>(m,n)</math>, our answer is <math>\boxed{005}</math>.
  
 
==See also==
 
==See also==

Latest revision as of 11:50, 30 July 2024

Problem

Find the number of ordered positive integer triplets $(a,b,c)$ such that $a$ evenly divides $b$, $b+1$ evenly divides $c$, and $c-a=10$.

Solution

Because $a|b$, let $b=an$, where $n$ is a positive integer. Because $c-a=10$, $c=a+10$, so $(b+1)|(a+10)$ and thus $(an+1)|(a+10)$. Now, let $a+10=m(an+1)$, where $m$ is another positive integer. Thus, $a=\tfrac{10-m}{mn-1}$. Because the ordered pair $(m,n)$ uniquely determines values of $a$, $b$, and $c$, the desired number of triples $(a,b,c)$ that fit the constaints of the problem equals the number of positive integer pairs $(m,n)$ that force $a=\tfrac{10-m}{mn-1}$ and, consequently, $b$ and $c$, to be positive integers. Starting with $n=1$, by listing out fractions of the form $\tfrac{10-m}{mn-1}$ and seeing if they simplify to positive integers, we see that the only possible values of $m$ are $2$ and $4$. Likewise, for $n=2$, $m$ must be $1$. For $n=4$, $m=1$, and for $n=10$, $m=1$. No other values of $n$ yield positive integer values of $m$. Thus, because there are $5$ ordered pairs $(m,n)$, our answer is $\boxed{005}$.

See also

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