Question

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

Answer 1

Dear student,
The given numbers are 520 and 468.

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.

Prime factorisation of 520 = 2 × 2 × 2 × 5 × 13

Prime factorisation of 468 = 2 × 2 × 3 × 3 × 13

LCM of 520 and 468 = 2 × 2 × 2 × 3 × 3 × 5 × 13 = 4680.(this is obtained by first taking terms common in both factorisation of 520 and 468(here 2×2×13 is common, hence we write, 2×2×13.) Now write the remaining numbers
i.e., 2x5 in 520 and 3x3 in 468).
Smallest number which when increased by 17 is exactly divisible by both 520 and 468 = 4680-17 = 4663.


Thank you.