Math, asked by vinod8068, 1 year ago

What is the smallest number that when divided by 3556 and 91 leaves remainder of 7 in each case?

Answers

Answered by mn121
2

3556 = 2*2*7*127

91 = 7*13

LCM = 2*2*7*127*13 = 46228

∴ Required number =  46228 + 7 = 46235

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