Q12 of 124 Page 1

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

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


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