Faulty Factorials

Relevant wiki: Factorials

We can rewrite the RHS of these equations as:

$\begin{array} { l l r } 1! \times 3! & = 3! & \{1\} \\ 1! \times 3! \times 5! & = 6! & \{2\} \\ 1! \times 3! \times 5! \times 7! & = 10! & \{3\} \\ 1! \times 3! \times 5! \times 7! \times 9! & = 15! \quad\quad & \{4\} \\ \end{array}$

The first equation is trivially true because $LHS = 1! \times 3! = 1 \times3! = 3! =RHS$ .

For the second equation, $LHS = 1!\times3!\times5! = 3!\times5! = (1\times2\times3)\times 5! = 6\times5! = 6!$ , as we applied the property of $(n+1)! = (n+1) \times n!$ . So the second equation is true as well.

Since we just know that the second equation is true, then $1! \times3!\times5! = 6!$ , and so $6!\times7! = 720\times7! = 10\times72 \times 7! = 10\times9\times8\times7! = 10!$ . So the third equation is true as well.

Lastly, since we just know that the second equation is true, then $1! \times3!\times5!\times7! = 10!$ , and so we want to determine whether $10!\times9! = 15!$ is true or not. But looking at the RHS, we can see that 13 (a prime number) divides $15!$ , but 13 does not divide any of the numbers $10!, 9!$ , so the fourth equation is not true.

There is an approach without much calculating. You have to look at the prime factorization. The prime factorization of any factorial of any integer n contains all the prime numbers that are smaller or equal to n. In the fourth equation on the RHS 15! has the prime factors 11 and 13. But on the LHS, there is no factor 11 or 13 because all the factorials that are summed up are smaller than 11 (and 13). And since prime numbers cannot be factorized by integers that are smaller than itself, {4} must be the solution. Of course I didn't prove that the other equations are valid but that's not necessary since its asked to find THE wrong equation and if I found one, then it has to be THE one.