Math, asked by kishankr, 1 year ago

what is the smallest number that when divided by 35 ,56 and 91 leaves remainder of 7 in each case

Answers

Answered by Anonymous
9
The smallest number which when divided by 35, 56 and 91 = LCM of 35, 56 and 91

35 = 5 x 7
56 = 2 x 2 x 2 x 7
91 = 7 x 13

LCM = 7 x 5 x 2 x 2 x 2 x 13 = 3640

The smallest number that when divided by 35, 56, 91 leaves a remainder 7 in each case = 3640 + 7 = 3647.

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