Question

The sum of a two-digit number and the number formed by reversing the order of digits is $66.$ If the two digits differ by $2,$ find the number. How many such numbers are there?

Answer 1

Let the digit at one's place be ${}^{\prime }{x}^{\prime }.$

Then, the digit at ten's place be $(x-2).$ [$\because$ Digits differ by $2$]

The original two digit number $=10(x-2)+x=10x-20+x=11x-20$

When the digits are reversed, then digit at ones place $=(x-2)$

And, the digit at tens place $=x$

So, the reversed number $=10x+(x-2)=11x-2$

It is given that the sum of the original two digit number and the reversed number is $66.$

$\therefore (11x-20)+(11x-2)=66$

$\Rightarrow 22x-22=66$

$\Rightarrow 22x=66-22$

$\Rightarrow 22x=88$

$\Rightarrow x=\frac{88}{22}=4$

So, digit at ones place $=x=4$

and, the digit at tens place $=x-2=4-2=2$

Hence, the original two digit number is $24.$

There will be two such numbers i.e, one will be $24$ and the other one will be $42.$