Question

What is the smallest number that, when divide by 35, 56 and 91 leaves remainders of 7 in each case?

Answer 1

We have to find the prime factorisation of 35, 56, and 91.

Prime factorisation of 35 = 5 × 7
Prime factorisation of 56 = 2 × 2 × 2 × 7
Prime factorisation of 91 = 7 × 13
∴ Required LCM = 2 × 2 × 2 × 5 × 7 × 13 = 3,640

Thus, 3,640 is the smallest number exactly divisible by 35, 56, and 91.
To get the remainder as 7:
Smallest number = 3,640 + 7 = 3,647
Thus, the required number is 3,647.