Math, asked by prasathharish5, 4 months ago

find the least number of five digits which when divided by 52,56,78 and 91 leaves no remainder​

Answers

Answered by mvandanamishra542
0

Answer:

The smallest number which when divided by 35, 56 and 91 = LCM(35,56,91)

35=5×7

56=2

3

×7

91=7×13

LCM=2

3

×5×7×13=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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