What's the value of two digits from a certain number composed by six digits when it comes from a prime?

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The problem is as follows:

The number $\overline{ab0ab0}$ is the result of the product between two consecutive prime numbers. Assuming the $0$ in the number represents zero. Find the value of $\overline{ab}$

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&38\\ 2.&85\\ 3.&51\\ 4.&34\\ 4.&17\\ \end{array}$

In my attempt to solve this problem I was only able to spot this by decomposing the six number digit as follows using base-10.

$\overline{ab0ab0}=10^5a+10^4b+10^3 \times 0+10^2a+10b+0=100100a+10010b$

Then I could note this can be rearranged as follows:

$10010(10a+b)$

Upon performing a canonical decomposition in the number I could spot that:

$10010= 2 \times 5 \times 7 \times 11 \times 13$

Had this question asked all those I would have landed in the answer. But that's where I'm stuck. Can someone help me from here?.

Am I heading in the right direction with this approach?. It would help me a lot if someone could help me from where I had came stuck.

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The problem makes some sense if we drop the word "two" and ask for just "a product of consecutive prime numbers". In such a case, having derived

$10010=2×5×7×11×13$

we observe that $3$ is missing. We multiply that in to get $30030$, which does not qualify because that would require the initial digit $a$ to be zero. To remove that blockage, try introducing the next prime factor, $17$, to get

$2×3×5×7×11×13×17=\color{blue}{51}0\color{blue}{51}0$.