Here is a fact everybody meets at school without noticing it is surprising. Take 10 factorial and divide it by 5 factorial times 5 factorial. The answer is 252, a whole number. Nothing in the arithmetic guarantees that: you are dividing one enormous number by another and getting no remainder. It works because the quotient counts something, namely the ways of choosing 5 things from 10, and a count cannot be a fraction.
Shift the terms slightly and the guarantee evaporates. Erdos problem 728 asks about this shape: when does a! b! divide n! k!, where k is a + b - n?
What makes it hard
If a + b is exactly n then k is 0, the question reduces to whether a! b! divides n!, and the answer is always yes because the quotient is a binomial coefficient. The interesting case is when a and b overshoot, so k is positive and both factorials on the bottom are large.
Erdos asked whether there are infinitely many triples where the divisibility holds while a and b each stay at least a fixed fraction of n. Keeping a and b large is easy. Making the divisibility work is easy if you let a and b be tiny. Doing both at once is the problem, and it stood open for decades.
You cannot explore it by computing factorials, which overflow immediately: 100 factorial already has 158 digits. The practical tool is Legendre's formula, which gives the exponent of a prime p in m factorial as floor(m/p) plus floor(m/p squared) plus floor(m/p cubed) and so on. The division works exactly when, for every prime, the exponent available is at least the exponent needed. The factorial divisibility calculator runs that test and reports the first prime that fails, which is almost always a small one, because small primes accumulate the largest exponents and so have the least room.
January 2026
On 4 January 2026, GPT-5.2 Pro produced a proof, in a session operated by Kevin Barreto. The argument was then formalised in Lean using Harmonic's Aristotle system, and a write-up was posted to arXiv. The result establishes a logarithmic-gap phenomenon: for suitable constants, there are infinitely many triples with a and b both between a fixed fraction of n and the same fraction below n for which the divisibility holds.
It was widely described as the first Erdos problem resolved more or less autonomously by an AI system, and Terence Tao characterised it as a real improvement in capability. One caveat is part of the record rather than a quibble: the informal problem statement was ambiguous, and the version that was proved had to be pinned down with stricter conditions than the original wording carried. That is a recurring feature of this whole episode, and it is the same issue that makes formalisation the bottleneck described in the piece on AlphaProof Nexus.
Trying it yourself
Set n to 100 and a and b to 50 and the divisibility holds, because a + b is exactly n and the quotient is the 30-digit binomial coefficient C(100,50). Raise either one and it fails immediately. Working through a few cases gives a feel for how rare the successes are, which is the part of the problem that is hard to convey from the statement alone.
For the other direction the same year took, where a machine produced a counterexample rather than a proof, see the unit distance conjecture.