Question

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

Answer 1

The smallest number which when increased by 17 is exactly divisible by both 468 and 520 is obtained by subtracting 17 from the LCM of 468 and 520.

Prime factorisation of 468 and 520 is:

468 = 2² × 3² × 13

520 = 2³ × 5 × 13

LCM = product of greatest power of each prime factor involved in the numbers = 2² × 3² × 5 × 13 = 4680

The required number is 4680 – 17 = 4663.

Hence, the smallest number which when increased by 17 is exactly divisible by both 468 and 520 is 4663.