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

Answer:

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 factorization of 468 and 520:

468 = 22 × 32 × 13

520 = 23 × 5 × 13

LCM = 22 × 32 × 5 × 13

LCM = 4680

The smallest number is 4680 – 17 => 4663.