My teacher gave me this tricky problem

please someone answer this question and write the solution as well.

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 (1) 8 (2) 10 (3)13 (4) 15 please reply friends.

#NumberTheory

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Comments

The $6-digit$ $number$ is in the form of $\color{#D61F06}{\overline{ababab}}$ .

Therefore ,

$\Rightarrow \color{#D61F06}{\overline{ababab}}$

$\Rightarrow 10^{5}a+10^{4}b+10^{3}a+10^{2}b+10a+b$

$\Rightarrow 101010a+10101b$

$\Rightarrow 10101(10a+b)$

$\Rightarrow 3×7×13×37×(10a+b)$

Now by observing the above line , we can say that $(10a+b)$ should have two prime factors other than $3,7,13$ and $37$ .

Also when we see $(10a+b)$ , we find that it is just a two digit number.

Now finding the number of numbers from $10-100$ having two prime factors.

$Case I$ :

Taking $2$ as one of the prime factors , we can say the upper limit as $2×50=100$ which means that the other prime number should be $2<P<50$ as we discussed earlier that the number should be of $2$ digits.

Therefore , number of prime numbers $2<P<50$ is $14$ but we don't have to count $3,7,13$ and $37$ .Thus , there are $\huge 10$ such numbers.

$Case II$ :

Now taking $5$ as one of the prime factors , the upper limit is $5×20=100$ . By this we can say that the another prime number should be $5<P<20$ .Now there are $5$ prime numbers $5<P<20$ but $7$ and $13$ are not to be considered. Therefore , there are $\huge 3$ such numbers.

Only these are the possibilities.

Therefore , Answer= $\huge \boxed{13}$