Consider numbers of the form \begin{align} p\equiv p_{a,b,n}&=a\times10^{n+1+\lfloor \log_{10}(b) \rfloor}+b\\ &=a\, \underbrace{0\cdots 0}_\text{$n$ zeros} \,b, \end{align} where $a,b,n\in\mathbb{N}$. For any $a,b$ (with $b$ odd and not ending in $5$), are there infinitely prime numbers of such form?
I've computed a few simple cases, for $n\leq 2000$, \begin{align} a= 1,b=1 &\Rightarrow n\in\{ 1 \}\\ a= 2,b= 1&\Rightarrow n\in\{ \}\\ a= 3,b= 1&\Rightarrow n\in\{2, 6, 9, 27, 35, 66, 80, 146, 482, 642, 1019, 1899 \}\\ a= 4,b= 1&\Rightarrow n\in\{ 1, 2, 12, 228, 241, 308, 956, 1472, 1493 \}\\ a= 9,b=1 &\Rightarrow n\in\{ 2, 3, 4, 8, 21, 26, 35, 56, 61, 77, 200, 536, 695, 789, 904, 1037 \}\\ a=1 ,b= 3&\Rightarrow n\in\{ 1, 4, 5, 10, 16, 17, 38, 55, 100, 104, 106, 122, 412, 425 \}\\ a=3 ,b=3 &\Rightarrow n\in\{ \}\\ a=4 ,b=3 &\Rightarrow n\in\{ 2, 6, 9, 39, 418, 448, 1736 \}\\ a=1 ,b=7 &\Rightarrow n\in\{ 1, 3, 7, 8, 23, 59, 109, 133, 221, 411, 699, 998, 1382 \}\\ a=2 ,b=7 &\Rightarrow n\in\{ \}\\ a= 4,b=7 &\Rightarrow n\in\{ 2, 8, 38 \}\\ a= 5,b=7 &\Rightarrow n\in\{ \}\\ a= 8,b=7 &\Rightarrow n\in\{ \}\\ a= 1,b= 9&\Rightarrow n\in\{ 1, 2, 3, 8, 17, 21, 44, 48, 55, 68, 145, 201, 271 \}\\ a=2 ,b=9 &\Rightarrow n\in\{ 4, 24, 454, 760 \}\\ a=4 ,b=9 &\Rightarrow n\in\{ 1, 3, 4, 7, 8, 27, 190, 195 \} \end{align} Some of these cases are relatively easy to show that there are not infinitely many such primes (for example $30\cdots 0 3$ is always a multiple of $10\cdots 01$), but some are rather intriguing, like $40\cdots 07$. Are these finite?
Any ideas or insights are appreciated.