Question

A two-digit number is such that the sum of the digits is 10. When 18 is subtracted from the number, the digits of new number formed become same. What is the number?

Answer 1

Let the digit of ones place of the two-digit number be `x` and that of tens place be `y`.

So the number is `10y+x`

By the condition of the problem we have

`y+x=10........[1]`

Again the nubmer obtained subtracting `18` from the two-digit number is

`10y+x-18`

`=10(y-2)+(x+2)`

This reveals that the new number has `(y-2)` in tens place and `(x+2)` in ones place. So by the given condition

`y-2=x+2`

`=>y-x=4.....[2]`

Adding [1] and [2] we have

`2y=14`

`=>y=7`

Inserting `y=7` in [1] we get

`7+x=10`

`=>x=3`

So the number is `10y+x=10*7+3=73`

Alternatve approach

Let the two-digit number obtained after subtracting 18 from the original two-digit number has got digit `x` both in ones place and tens place.

So this numer will be `10x+x=11x`

and the original number was

`11x+18=10(x+2)+(x-2)`

So the original number has `(x+2)` in tens place and `(x-2)` in ones place.

So by the given condition we can write

`(x+2)+(x-2)=10`

`=>x=5`

Hence the original number was

`11x+18=11*5+18=73`