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This question already has an answer here:

Yesterday, when I was playing with numbers, I was surprised to know the following relation:

$$10!=6!\times 7!.$$

Then, I've been looking for the other solutions, but I'm facing difficulty. Then, here are my questions.

Question 1 : Does there exist the other non-trivial solution?

Question 2 : Can we get all solutions? In addition to this, can we prove that they are all solutions?

I don't know if these questions are famous. Also, I'm afraid that these questions might be very easy to solve. Anyway, I need your help.

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marked as duplicate by Andreas Caranti, Gerry Myerson, Peter Taylor, user1337, Ayman Hourieh Sep 15 '13 at 13:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Hint: Notice that $7$ is the largest prime $\leq 10$.

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    $\begingroup$ How exactly does this contribute to solving $a!=b!c!$. $\endgroup$ – Gerry Myerson Sep 15 '13 at 12:37
  • $\begingroup$ that is one of the reasons for $10\times9\times8\times7$ to be equal to $7!$. We need to search for examples so that $b$ or $c$ is the largest prime lesser than $a$, that somehow narrows down the search. $\endgroup$ – Abishanka Saha Sep 15 '13 at 12:43
  • $\begingroup$ @AbishankaSaha: Thank you very much! $\endgroup$ – mathlove Sep 15 '13 at 14:59

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