Question

find the smallest no. which leaves remainder 7 when divided by 35, 56 and 91 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

×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.

Answer 2

Answer:

now add 7 in that

Step-by-step explanation:

plz mark me as Brainliest