This question is taken from real numbers chapter in which we have given three numbers 35, 56 and 91 and we have been asked to find the smallest number which when divided by the above number leaves remainder 7 in each case.

Kindly give me a detailed solution of this question

RS Aggarwal, Class 10, chapter 1B, question no 10

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

35=5×7

56=2×2×2×7

91=7×13

LCM =7×5×2×2×2×13=3640

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