Find the least number which leaves a remainder of 8 when divided by 20, 24 or 45, and which is also exactly divisible by 13.

LCM(20, 24, 45) = $2^3\times 3^2\times$ 5 = 360, so the number has the form $360k+8$. Testing $k=1$, 2, 3 gives 368, 728 and 1088, and only $728=13\times 56$ is divisible by 13. Answering 368 stops at the first candidate and ignores the divisibility-by-13 condition; 360 is the LCM itself, with the remainder left off.