Find the smallest number which, when divided by 9, 10 and 21, leaves a remainder of 4 in each case.

The number is 4 more than a common multiple of 9, 10 and 21. LCM(9, 10, 21) = $2\times 3^2\times 5\times 7=630$, so the smallest such number is 630 + 4 = 634. Answering 630 forgets to add the remainder, 626 subtracts it instead of adding, and 1264 uses the second common multiple.