Question

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

Answer 1

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.

Answer 2

Given: Three numbers - 35, 56 and 91

To find: The least number which when divided by the given numbers leaves remainder 7 in each case

Solution: To find such a number, first we need to find the least number which is exactly divisible by the given numbers. That required number is called Least Common Multiple.

Using Prime Factorization method:

35 = 5 × 7

56 = 2 × 2 × 2 ×7

91 = 7 × 13

Now we need to check the maximum frequency of all the numbers which are factors of 35, 56 and 91.

5 occurs maximum 1 time, 2 occurs maximum 3 times, 7 occurs maximum 1 time and 13 also occurs maximum one time.

L.C.M. = 2 × 2 × 2 × 5 × 7 × 13 = 3640

3,640 is the number which is exactly divisible by given numbers. Now, the required number is obtained by adding 7 to 3640 i.e., 3640 + 7, which is 3647.

Hence, 3647 is the number that will leave remainder 7 in each case on being divided by the given numbers.