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
If we look at this modulo 12, we see that the LHS is congruent to and the RHS is congruent to 1. Therefore . 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.