in base 10, two-digit prime n is 45 more than the number formed by the reversing the digits of n. find all possible values of n.

Question

anonymous · Answer

Let $n_1$ and $n_2$ be the numbers in the 10s and 1s digit, respectively; then,
$$n=10n_1+n_2$$
The given info gives us the following equation:
$$\begin{align*}10n_1+n_2&=45+10n_2+n_1\
10n_1+n_2&=10(n_2+4)+(n_1+5)
\end{align*}$$
Mathcing up the 10s and 1s digits of these numbers, you have
$$\begin{cases}n_1=n_2+4\n_2=n_1+5\end{cases}$$
I'm afraid there aren't any solutions to this system, but it's possible my reasoning is flawed. I just thought I'd try my hand at it.

ganeshie8 · Answer

$$
\begin{cases}n_1=n_2+4\n_2=n_1+5\end{cases} 

$$

is valid only when $n_1 \le  4$

ganeshie8 · Answer

for $n \gt 4$ we can use :
$$
\begin{cases}n_1=n_2+5\n_2=n_1-5\end{cases} 

$$
 $\implies  n_1 - n_2 = 5$ 

solving in single digit positive integers gives two solutions :  $61,  83$