Answer to Question #119510 in Physics for Hussain Ousman Darboe

Question

A number of two digits is increased by 9. When the digits are reversed, the sum of the digits is 5. Find the number?

Answer 1

The original condition is:

The sum of the digits of a number lying between 10 and 100 is 5. When the digits are reversed, the number is increased by 9. Find the number.

Assume that the initial number is

$$10a+b.$$

If the digits are reversed, the number of two digits is increased by 9:

$$10b+a=10a+b+9,\\ 9b-9a=9,\\ b-a=1.$$

The sum of the digits is 5:

$$a+b=5.$$

Therefore, we have two simple equation that form a system. The solution of the system

$$b-a=1\\ a+b=5$$

is

$$a=2\\ b=3.$$