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 $\boxed{005}$.

See also