Question

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

Answer 1

The smallest number which when increased by 17 is exactly divisible by both 520 and 468 is obtained by subtracting 17 from the LCM of 520 and 468. Smallest number which when increased by 17 is exactly divisible by both 520 and 468 =4680– 17 = 4663. So, least number divisible by both 520 and 468 is 4680.

Answer 2

LCM of 520 and 468 = 4680

So, smallest no. when incresed by 17 divisible by the given numbers
is = 4680-17 = 4663