Is every factorial divisible by its sum of digits?

Denote by $\Sigma_d(t)$ the sum of digits in the decimal representation of the number $t$.

Prove / disprove:

$$\forall n\in \mathbb N:\ \ \Sigma_d (n!) | n!$$

• $\Sigma_d (n!) = O(n \ln (n))$ grows faster than $n$, so there's no clear reason to think it's asymptotically true. Have you looked for counterexamples? Jul 27, 2014 at 9:57
• Table[Mod[n!, Plus @@ IntegerDigits[n!]], {n, 1, 500}] Jul 27, 2014 at 10:15
• what about sum of binary digits? (following the usual observation that there's nothing special about 10) Jul 27, 2014 at 17:41
• @Mitch, I wrote a program to generate the first counterexamples in arbitrary bases. For bases 2 through 30, the first counterexamples are 10, 43, 86, 87, 188, 156, 291, 364, 432, 410, 7, 510, 4, 4, 4, 813, 4, 1079, 4, 1900, 6, 10, 6, 2330, 2147, 5, 3463, 2401 and 7 respectively. Jul 28, 2014 at 10:04
• @G.H.Faust: Add it to OEIS!! Jul 28, 2014 at 12:06

It's not true. The first counterexample is for $n = 432$. The sum of the digits in $432!$ is 3897, which you can see using Wolfram Alpha. But the prime factorisation of 3897 is $3^2 \times 433$, so $432!$ cannot be divisible by its sum of digits.
• The fact that the first counterexample has $n + 1$ as its only large prime factor is somewhat remarkable. It's like the hole-in-one of counterexamples. Jul 27, 2014 at 13:19
• Curious that $432, 532, 632$ are each counterexamples (though not $732$) Jul 27, 2014 at 20:34