ICTM Regional 2009 Divsion AA

If $x$ divided into 3587 leaves a remainder of 12, then we can write this as:

$3587 \equiv 12 \text{ (mod x)}$

$3575 \equiv 0 \text{(mod x), or x is a factor of}\quad 3575$

Note that $3575 = 5^2 \times 11 \times 13$ and its factors are:

$1, 5, 11, 13, 25, 55, 65, 143, 275, 325, 715, 3575$

However since neither $1, 5$ nor $11$ can leave a remainder of 12 when divided 3587 because they are less than 12, they cannot be counted.

$3587 \div 1 \Rightarrow \text{remainder} \quad 0$

$3587 \div 5 \Rightarrow \text{remainder} \quad 2$

$3587 \div 11 \Rightarrow \text{remainder} \quad 1$

So in total there are $\boxed{9}$ distinct positive integers that leave a remainder of 12 when divided into 3587:

$13, 25, 55, 65, 143, 275, 325, 715, 3575$