Question

The sum of the digits of a two-digit number is 9. If the difference of the original
number and the number obtained by reversing the digits is 63, find the original
number.

Answer 1

In context to question asked,

We have to determine the value of original number.

As per question,

We have,

The sum of the digits of a two-digit number is 9.

The difference of the original  number and the number obtained by reversing the digits is 63

As we know that,

The general expression of a two digit number is $10x+y$ where "x" is the ten's digit number and y is unit's digits number.

So, let the value of original number is $10x +y$

As when the digits of the number are reversed, means the digits of unit place and ten's place are interchanged.

So, we can rewrite the reversed number as $10y+x$

So, from question,

We can rewrite it as,

$[10x+y] -[10y-x] = 63\\=>10x+y-10y-x = 63\\=>10x-x+y-10y=60\\=>9x-9y=63\\=>9(x-y)=63\\=>x-y = \frac{63}{9}\\=>x-y=7-------------(1)$

As the sum of digits of the two digits number is 9.

So, $x+y=9-----------(2)$

So, from (1) and 92),

We will get,

$x-y=7 -----(1)\\x+y=9-----(2)\\Adding\:(1) \:and\:(2),\\=>2x =16\\=>x = \frac{16}{2}=8\\=>x = 8\\=>y = 9-8 = 1\\=>Number = 10x+y =10\times 8 +1=80+1=81$

Hence, value of actual number will be 81.