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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.

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    $\begingroup$ 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. $\endgroup$ – rogerl Apr 18 at 21:11
  • $\begingroup$ $b$ can't be $2$, because $10c+b$ would be even, $b$ can't be $5$ too $\endgroup$ – M. Di Apr 18 at 21:13
  • $\begingroup$ $2$ 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 $\endgroup$ – WaveX Apr 18 at 21:15
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$b$ 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 $a=6, b=7, c=9$.

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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.

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