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Question
What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case?
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Solution
35 = 5 × 7
56 = 2 × 2 × 2 × 7
91 = 7 × 13
L.C.M. of 35, 56, 91 = 5 × 7 × 2 × 2 × 2 × 13 = 3640
∴ Required number = 3647 which leaves remainder 7 in each case.
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