Find the smallest number, divisible by 13, such that the remainder is 1 when divided by 4,6 or 9.
The number is congruent to 1 mod 4, 1 mod 6, 1 mod 9 so it must be congruent to 1 mod 36 (36 being lcm(4,6,9)). So we want to find integer solutions (a,b) to
$\displaystyle 13a = 36b + 1$
If we look at this modulo 12, we see that the LHS is congruent to $\displaystyle a$ and the RHS is congruent to 1. Therefore $\displaystyle a \equiv 1 (\mod 12)$. a = 1, a = 13 do not work, but a = 25 works. Hence the smallest possible positive multiple of 13 that works is 13*25, or 325.