Question

What is the least multiple of 11, which when divided by 3, 4, 5, 6 and 7, leaves the remainder 2, 3, 4, 5 and 6 respectively ?​

Answer 1

Answer:

119

Suppose we are given that a number when divided by x, y, and z, leaves a remainder of a, b, and c; then the number will be of the format of

LCM(x,y,z)*n + constant

The key in these questions is finding out the value of 'constant'. If all of them leave the same remainder 'r', constant = r. It can also be looked at as the smallest number satisfying the given property.

In this question, we are given

Remainder from 2 is 1

Remainder from 3 is 2

Remainder from 4 is 3

Remainder from 5 is 4

Remainder from 6 is 5

If you look at the negative remainders

Remainder from 2 is -1

Remainder from 3 is -1

Remainder from 4 is -1

Remainder from 5 is -1

Remainder from 6 is -1

So, the number N = LCM(2,3,4,5,6)*n - 1 = 60n - 1

So, any number which is of the format of 60n - 1 will satisfy the given conditions.

Some such numbers are 59, 119, 179, 239....

We want the one which is divisible by 7.

59 is not divisible by 7

119 is divisible by 7 and our answer