Find two or more different positive integers the product of whose factorials is a perfect cube.

How small can the largest of these be?

How few can they be?


2 Answers 2


How small can the largest of these be?

Nine, in order to get three factors of 7. Otherwise it would have to be less than 7, but then there couldn't be any factors of 3 or 5.

Proof positive:

$4! \times 7! \times 8! \times 9! = 2^{21} \times 3^9 \times 5^3 \times 7^3 = (2^7 \times 3^3 \times 5 \times 7)^3$

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    $\begingroup$ Just beat me! Here's another one: $5!23!24!25! = 2^{66}×3^{30}×5^{15}×7^9×11^6×13^3×17^3×19^3×23^3$ $\endgroup$
    – RobPratt
    Jul 3, 2022 at 16:25
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    $\begingroup$ Confirmed by brute force that this is the only 4-factor solution with factors of 20! or less. $\endgroup$
    – dan04
    Jul 5, 2022 at 20:38

Possibly-partial answer (if "how few can they be?" means how few absolutely, rather than how few conditional on making the largest as small as possible, I have not answered that):

[There is already an answer finding the same minimax example as mine. I didn't post this earlier because I was trying to make progress on the "minimum number of factorials" question. In any case, this answer differs from that one by having more explanation, and by handwaving a bit about the question that one doesn't address.]

If $p$ is any prime between $n/2$ and $n$ exclusive then $n!$ is divisible by $p$ but not by $p^2$. So if we let $p$ be, say, the largest prime $\leq$ all of our integers, then (by a famous number-theoretic result known as Bertrand's Postulate, so called of course because it was first proved by Chebyshev :-), which says that for any integer $n>1$ there is a prime number between $n$ and $2n$ exclusive) the number of factors of $p$ dividing the product of the factorials is just the number of integers $\geq p$. So this must be at least 3, and in particular the total number of integers whose factorials we're taking must be at least 3.

[Credit where due: earlier versions of the paragraph above contained an idiotic mistake; thanks to Greg Martin for spotting it.]

Obviously we can make the product a cube by taking three equal integers, or rather we could if the question didn't specifically say "three or more different positive integers". But maybe we can take three nearby integers, with no primes lying between the smallest and largest, and have the "extra" factors work out.

Well, the first pair of primes differing by 3 or more is 7 and 11. So maybe we can take three of 7!, 8!, 9!, 10! and get a perfect cube? Dividing out the $7!^3$ factor we will obviously get, can we take three of the numbers 1, 8, 72, 720 and make the product a cube? Alas, no. If we include 720 then we get a lone factor of 5. If not, our numbers are 1, 8, 720; the first two are cubes, the third isn't. Fail.

But maybe we can combine some of these with some smaller factorials? We need their product with either 72 or 72.720 to be a cube; equivalently, discarding factors of 8 and 27, we need their product with either 9 or 30 to be a cube. And yes, we can do this. 4! = 24 is three times a cube.

So, we can do it with four factorials: 4! 7! 8! 9! is a cube. And it's clear from the above that 9 is the smallest possible largest integer, and that with 9 as largest integer we can't use fewer than four factorials.

Can we do it with three factorials, if we allow larger integers? If so, they must all be factorials of numbers in $[p,q)$ where $p,q$ are consecutive primes. This is impossible if $q=p+2$. If $q=p+4$ then we need the product of three of $1$, $p+1$, $(p+1)(p+2)$, $(p+1)(p+2)(p+3)$ to be a cube; a bit of computer searching finds no examples of reasonable size. Indeed, a bit of computer searching finds no examples of reasonable size whatever the gap between $p,q$. (I have tried up to about 150000. My code is very simple-minded and inefficient.) Perhaps the thing is impossible for products of only three factorials? I don't immediately see a plausible way to try to prove it.

Well, let's see. Suppose we have three "nearby" factorials a! b! c!, so the product is a cube times $(b!/a!)(c!/a!)$, which is to say a cube times $\bigl[(a+1)\cdots(b)\bigr]^2\bigl[(b+1)\cdots(c)\bigr]$. All these factors are close to one another, which means they are "nearly coprime". So e.g. suppose one of these numbers from $a+1$ to $c$ is divisible by a prime $>c-a$; then none of the others can be, which means that product cannot possibly be a cube. So we need a run of numbers divisible only by rather small primes. That might well be impossible. It might be impossible but inaccessible to proof because it just amounts to "a very low-probability coincidence happens not to occur" (analogy: Fermat primes bigger than $2^{2^4}+1$). It might actually happen "randomly" at some annoyingly large $n$ (maybe somewhere where this an unusually long gap between consecutive primes). (Of course nothing to do with the prime numbers is really random, but in many respects they behave as if they are...)

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    $\begingroup$ "If $p$ is any prime $\geq\sqrt n$ then $n!$ is divisible by $p$ but not by $p^2$" is incorrect (consider $p=3$ and $n=7$ for example). The $\ge\sqrt n$ must be replaced by $>\frac n2$ (and also the assumption $p\le n$ must be included). Moreover, some justification is needed why there exists such a prime for each $n$—for the corrected statement it is Bertrand's postulate. $\endgroup$ Jul 3, 2022 at 23:57
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    $\begingroup$ Whoops! Of course I meant n/2 not sqrt(n); will fix. I don't think I assumed that there does exist such a prime for each n (though, as you say, that is in fact true): where do you think I did? $\endgroup$
    – Gareth McCaughan
    Jul 4, 2022 at 21:00
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    $\begingroup$ Oh, duh, indeed I am assuming it and will make an appropriate adjustment to my answer. $\endgroup$
    – Gareth McCaughan
    Jul 4, 2022 at 21:01

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