Product of Factorials into a Factorial

Number Theory Level 2

Given a positive integer $n$ , does there always exists integers $a_1,a_2,\ldots a_n, b$ greater than $1$ such that $a_1!a_2!\ldots a_n!=b!$

No Yes

1 solution

Chris Lewis
Aug 4, 2020

Some nice small examples are $10!=6!\cdot 7!=3!\cdot 5!\cdot 7!$ .

In general, we have $\left(2^{n-1}\right)!=(2!)^{n-1} \cdot \left(2^{n-2}\right)!$

so the answer is yes . For example, with $n=5$ , we have $16!=2! \cdot 2! \cdot 2! \cdot 2! \cdot 15!$ .

Of course, we don't need to choose powers of $2$ ; any power of a factorial will work in the same way (eg $216!=3! \cdot 3! \cdot 3! \cdot 215!$ ), but these solutions aren't fundamentally different.

Whether there are solutions for every $n$ if we insist the $a_i$ are distinct, I don't know. Bonus question?

I found out that $3!! = 6!$

Yajat Shamji - 10 months, 1 week ago

There are solutions for every $n$ where all the $a_i$ are distinct. The following equations give the first few examples:

$3! \cdot 5! = 6!$

$3! \cdot 5! \cdot 719! = 720!$

$3! \cdot 5! \cdot 719! \cdot (720! - 1)! = (720!)!$

Jon Haussmann - 10 months, 1 week ago

Of course! Very nice. And you could start those with any factorial (eg $4! \cdot 23!=24!$ etc).

Chris Lewis - 10 months, 1 week ago

Product of Factorials into a Factorial

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