Question

What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case?

Answer 1

SOLUTION :  

GIVEN : Three numbers 35, 56 and 91.

Remainder = 7

To find the smallest number, first, we need to find the LCM of (35, 56 and 91)  and then add remainder 7 in the LCM.  

The prime factors of 35, 56 and 91 are :  

35 = 5 x 7

56 = 2³ x 7

91 = 13 x 7

L.C.M of 35, 56 and 91 = 2³ x 7  x 5 x 13  

[LCM of two or more numbers = product of the greatest power of each prime factor involved in the numbers, with highest power.]

L.C.M of 35, 56 and 91 = 3640

3640 is the least number which exactly divides 28, 42 and 84 i.e. we will get a remainder = 0 in each case. But we want the smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case.

Therefore = 3640+ 7 = 3647

Hence, 3647 is smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case.

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Answer 2

L.C.M of 35 56 91 is

35 = 5 x 7

56 = 2 x 2 x 2 x 7

91 = 7 x 13

And the lcm is equals to 3640

Rrminder is 7 then

3640 + 7 = 3647