2013 Mock AIME I Problems/Problem 2

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 $\boxed{005}$ .

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2013 Mock AIME I Problems/Problem 2

Problem

Solution

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