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

Question

Philoid · Accepted Answer

Therefore the required smallest number would be the LCM of the numbers

Prime factors of 35 = 5 × 7

Prime factors of 56 = 2 × 2 × 2×7

Prime factors of 91 = 7 × 13

LCM of 35, 56 and 91 ⇒ 2 × 2 × 2 × 5 × 7×13 = 3640

Number = LCM + Remainder

⇒ 3640 + 7 = 3647