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.

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## Comments

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TopNewestThe \(6-digit\) \(number\) is in the form of \(\color{red}{\overline{ababab}}\).

Therefore ,

\(\Rightarrow \color{red}{\overline{ababab}}\)

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

\(\Rightarrow 101010a+10101b\)

\( \Rightarrow 10101(10a+b)\)

\(\Rightarrow 3×7×13×37×(10a+b)\)

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=\(\huge \boxed{13}\)Log in to reply

Nice!

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\(\huge Thanks!!\)

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Thanks Akshat for your amazing solution. You have proved the title to be wrong. I wish I could upvote your solution thousand times. And thank you Mohit for your answer.

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^_^

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Yes , its \(13\).

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Answer is 13

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you may be correct since i don't know the correct answer so please write the solution and thank you for your instant reply. i was not able to answer this question so i made a note.

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I think that you'll find my solution useful ^_^

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