# My teacher gave me this tricky problem

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.

3 years, 5 months ago

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The $$6-digit$$ $$number$$ 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=$$\huge \boxed{13}$$

- 3 years, 5 months ago

Nice!

Staff - 3 years, 5 months ago

$$\huge Thanks!!$$

- 3 years, 5 months ago

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.

- 3 years, 5 months ago

^_^

- 3 years, 5 months ago

- 3 years, 5 months ago

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.

- 3 years, 5 months ago

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

- 3 years, 5 months ago

Yes , its $$13$$.

- 3 years, 5 months ago