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The number of 6 -digit numbers of the form ababab (in base 10$)$ each of which is a product of exactly 6 distinct primes is
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13
$N=a b a b a b$
$\begin{array}{llll}1 < a \leq 9 & 0 < b \leq 9 & a, b \in \mathrm{l}\end{array}$
$N=10^{5} a+10^{4} b+10^{3} a+10^{2} b+10 a+b$
$=\left(10^{4}+10^{2}+1\right)(10 a+b)$
$=\left(10^{2}+10+1\right)\left(10^{2}-10+1\right)(10 a+b)$
$=3 \times 37 \times 13 \times 7(10 a+b)\quad ......\text {(1)}$
$\begin{array}{lcc}\text { then } & 10 \mathrm{a}+\mathrm{b} =\mathrm{P}_{1} \times \mathrm{P}_{2} & \mathrm{p}_{1}, \mathrm{p}_{2} \in \text { prime and } 10 \leq 10 \mathrm{a}+\mathrm{b} \leq 99 \\ \mathrm{a} & \mathrm{b} & 10 \mathrm{a}+\mathrm{b} \\ 1 & 0 & 10=2 \times 5 \\ 2 & 2 & 22=2 \times 11 \\ 3 & 4 & 34=2 \times 17 \\ 3 & 8 & 38=2 \times 19 \\ 4 & 6 & 46=2 \times 33 \\ 5 & 5 & 55=5 \times 11 \\ 5 & 8 & 58=2 \times 29 \\ 6 & 2 & 62=2 \times 31 \\ 7 & 4 & 74=2 \times 37 \\ 8 & 2 & 82=2 \times 41 \\ 8 & 5 & 85=5 \times 17 \\ 9 & 4 & 94=2 \times 47 \\ 9 & 5 & 95=5 \times 19\end{array}$
$\begin{array}{llll}1 < a \leq 9 & 0 < b \leq 9 & a, b \in \mathrm{l}\end{array}$
$N=10^{5} a+10^{4} b+10^{3} a+10^{2} b+10 a+b$
$=\left(10^{4}+10^{2}+1\right)(10 a+b)$
$=\left(10^{2}+10+1\right)\left(10^{2}-10+1\right)(10 a+b)$
$=3 \times 37 \times 13 \times 7(10 a+b)\quad ......\text {(1)}$
$\begin{array}{lcc}\text { then } & 10 \mathrm{a}+\mathrm{b} =\mathrm{P}_{1} \times \mathrm{P}_{2} & \mathrm{p}_{1}, \mathrm{p}_{2} \in \text { prime and } 10 \leq 10 \mathrm{a}+\mathrm{b} \leq 99 \\ \mathrm{a} & \mathrm{b} & 10 \mathrm{a}+\mathrm{b} \\ 1 & 0 & 10=2 \times 5 \\ 2 & 2 & 22=2 \times 11 \\ 3 & 4 & 34=2 \times 17 \\ 3 & 8 & 38=2 \times 19 \\ 4 & 6 & 46=2 \times 33 \\ 5 & 5 & 55=5 \times 11 \\ 5 & 8 & 58=2 \times 29 \\ 6 & 2 & 62=2 \times 31 \\ 7 & 4 & 74=2 \times 37 \\ 8 & 2 & 82=2 \times 41 \\ 8 & 5 & 85=5 \times 17 \\ 9 & 4 & 94=2 \times 47 \\ 9 & 5 & 95=5 \times 19\end{array}$
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