The difference of the digits of a two-digit number is 2. The sum of that two-digit number and the number obtained by interchanging the places of its digits is 132. Find the two-digit number(s).

Given:

The difference of the digits of a two-digit number is 2. The sum of that two-digit number and the number obtained by interchanging the places of its digits is 132 

 To do:

We have to find the number.

Solution:

Let the two-digit number be $10x+y$.

According to the question,

$x-y=2$ or $y-x=2$-----(i)

The number obtained on reversing the digits is $10y+x$.

$10y+x+(10x+y)=132$

$10y+y+x+10x=132$

$11y+11x=132$

$11(y+x)=132$

$y+x=12$

$y=12-x$

Substituting the value of $y$ in equation (i), we get,

$x-(12-x)=2$ or $12-x-x=2$

$2x-12=2$ or $2x=12-2$

$2x=12+2$ or $2x=10$

$2x=14$ or $x=5$

$x=7$ or $x=5$

This implies,

$y=12-7=5$ or $y=12-5=7$

$10x+y=10(7)+5=70+5=75$ or $10x+y=10(5)+7=50+7=57$

The required number is $75$ or $57$.  

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Updated on: 10-Oct-2022

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