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Question

What is the smallest number that when divided by 35 , 36 and 91 leaves remainder 7 in each case?


Solution

35=5×7

36=2^2 × 3^2

91=7×13

So,LCM(35,36,91)=2^2 × 3^2 ×5×7×13 =36×35×13=36×455=16380

Hence required number =16380+7=16387 

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