Question

Find the least number which when divided by 35, 56, and 91 leaves the same remainder
7 in each case

Answer 1

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

35=5×7

$56=2^{3} \times 7$

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.