# A 3 digit number abc, $N=b(10c+b)$ where $b$ and $(10c+b)$ are primes.

A three-digit number $$N$$ has first digit $$a$$ (not equal $$0$$), second digit $$b$$ and third digit $$c$$. $$N=b(10c+b)$$ where $$b$$ and $$(10c+b)$$ are primes. Find $$N$$.

$$N = 100a+10b+c$$ , then $$100a+10b+c = b(10c+b)$$

I don't know how to proceed.

• I think modified trial and error is the only way to go. What can $b$ be? There are only four possibilities: 2, 3, 5, 7, and you can eliminate two of those given that $10c+b$ must be prime as well. – rogerl Apr 18 at 21:11
• $b$ can't be $2$, because $10c+b$ would be even, $b$ can't be $5$ too – M. Di Apr 18 at 21:13
• $2$ and $5$ can only be removed once it has been shown that $c\neq 0$ and this in itself isn't that hard to show – WaveX Apr 18 at 21:15

$$b$$ is a single digit prime; $$\{2,3,5,7\}$$. But $$b$$ is also the last digit of a two digit prime $$10c+b$$, and no two digit prime can end in either $$2$$ or $$5$$. So $$b$$ is either $$3\ \text{or}\ 7$$. Since $$N=b(10c+b)$$, the last digit of $$N$$ is the last digit of $$b^2$$, which in either case is $$9$$. So $$10c+b$$ is either $$93$$ or $$97$$. Of those two, only $$97$$ is prime. So $$b$$ must be $$7$$. $$N=7\cdot 97=679$$ so $$a=6, b=7, c=9$$.
The only solution is $$N=679=7(97)$$ by a selective search. We know that $$b\in\{3,7\}$$ as $$10c+b$$ is prime so the search isn't that hard.