Question

What is the smallest number that both 33 and 39 divide leaving remainders of 5?

Answer 1

We have to find prime factorisation of 33 and 39.

Prime factorisation of 33 = 3 × 11
Prime factorisation of 39 = 3 × 13
∴ Required LCM = 3 × 11 × 13 = 429

Thus, 429 is the smallest number exactly divisible by 33 and 39.
To get the remainder as 5:
Smallest number = 429 + 5 = 434
Thus, the required number is 434.