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