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?

1 Answer
Apr 2, 2017

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