When a two digit number is divided by the number formed by exchanging the two digits, the quotient is equal to its remainder.

Question

This question is of number theory. I used the division algorithm also but to no avail.

I did it like $10a+b=(10b+a)(x)+x$ then firstly I found out $x$ as $10a+b/10b+a+1$ and in another attempt I did it like $a(10-x)+b(1-10x)=x$ where $x$ is the quotient and the remainder which are equal. But I am not able to go further.

Answer 1

Hint

If $\overline{ab}$ is the two digits number you’re looking for, the hypothesis writes as

$$10a+b = (10b+a) r +r$$

We have $a>b$ as otherwise we get a contradiction. If the exchange of $a,b$ provides a two digits number we need to have $a>b\ge 1$.

Therefore $\overline{ab} \lt 98$, $\overline{ab}\gt 12$ and $1 \le r \le 8$.

From there, one option is to study the cases by picking up each option for $r$ and solving the equation

$$a(10-r) - b(10r-1)= r$$

For example if $r=1$, we get $9(a-b)= 1$ which is impossible.

And so on...