Let \(N=10x+y\), so that \(x\) is the digits in the tens place and \(y\) is the digit in the ones place. (Clearly \(x
eq0\).)
\(10x+y=45+10y+x\implies 9x-9y=45\implies x-y=5\)
There are five possible two digits integers that satisfy this relation:
\(x=5,y=0\implies N=50\)
\(x=6,y=1\implies N=61\)
\(x=7,y=2\implies N=72\)
\(x=8,y=3\implies N=83\)
\(x=9,y=4\implies N=94\)
But the first, third, and fifth candidates are even, so they are not prime. The remaining are prime, however, so \(N=61\) or \(N=83\).
