Question

find the smallest number which when increased by 17 is exactly divisible by both 520 and 468 .plz exllain with method

Answer 1

Prime factors of 

520 = 2 × 2 × 2 × 5×13
468 = 2 × 2 × 3 × 3 × 13

LCM = 2 × 2 × 2 × 3 × 3 × 5 × 13 = 4680

So smallest no. = 4680-17
                         =4663

Answer 2

prime factors of 520 and 468 are:
468 = 2*2*2*5*13
568 = 2*2*3*3*13
now we take L.C.M
= 2*2*2*3*3*5*13
=4680
then smallest no. is = 4680-17
= 4663
4663 is the answer.