Question

A certain number when divided by a certain divisor leaves a remainder 20. When twice the number id divided by the same divisor then the remainder is 9. What is the 15th multiple of the divisor ?

Answer 1

`15^(th)` multiple of divisor is `465`

Explanation:

Let the number be `x` and divisor be `d`. As this leaves a remainder `20`, `d>20` and

`x=n_1d+20`, where `n_1` is quotient and a whole number

This also means `2x=2n_1d+40` .............(A)

As twice the number is `2x` and when this is divided by `d` remainder is `9`, we have

`2x=n_2d+9` .............(B)

from (A) and (B), we get `2n_1d+40=n_2d+9`

i.e. `(n_2-2n_1)d=40-9=31`

As remainder reduces after dividing `2x` by `d` as compared to dividing `x` by `d`, `d` must have gone one more time than twice `n_1` and we must have `n_2-2n_1=1`.

therefore `d=31`

and `15^(th)` multiple of divisor is `15xx31=465`