Question

Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468

Answer 1

The smallest number which is divisible by both 520 and 468 will be the lowest common multiple of 520 and 468.
L.C.M of 520 and 468,
520 = 2 * 2 * 2 * 5 * 13
468 = 2 * 2 * 3 * 3 * 13
So L.C.M will be 2 * 2 * 2 * 3 * 3 * 5 * 13 = 4680
But we have to find the number which, when increased by 17 is exactly divisible by 520 and 468, so we just subtract 17 from 4680 i.e. 4663 is our answer.