Question

find the smallest number which when increased by 17 is exactly divisble 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 x 2 x 2 x 5 x 13

468 = 2 x 2 x 3 x 3 x 13

So L.C.M = 2 x 2 x 2 x 3 x 3 x 5 x 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.
That is,
4680 - 17 = 4663

4663 is the answer.