Denote by $\Sigma_d(t)$ the sum of digits in the decimal representation of the number $t$.
Prove / disprove:
$$\forall n\in \mathbb N:\ \ \Sigma_d (n!) | n!$$
Denote by $\Sigma_d(t)$ the sum of digits in the decimal representation of the number $t$.
Prove / disprove:
$$\forall n\in \mathbb N:\ \ \Sigma_d (n!) | n!$$
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 list of counterexamples is sequence A066419 in the OEIS.