Given is a prime number p. Prove that the number p⋅(p2⋅pp−1−1p−1)!is divisible by p∏i=1(pi)!.
Problem
Source: Polish MO Second round 2024 P6
Tags: factorial, number theory
10.02.2024 23:23
Although it looks terrible at first sight, it is actually a very nice problem (as is of course to be expected from it being on the Polish olympiad).
11.02.2024 00:33
one can do the following:
11.02.2024 01:38
Oh, that's very cute!
11.03.2024 05:56
Note that as (a_1)! \cdots (a_n)! divides (a_1 + \cdots + a_n)! we have that both: a = \frac{(p^2 + \cdots + p^p)!}{(p^2)! \cdots (p^p)!}And b = \frac{(p - 1 + p^2 - 1 + \cdots + p^p -1)!}{(p-1)! (p^2 - 1)! \cdots (p^p - 1)!} = \frac{(p^2 + \cdots + p^p)!}{(p-1)! (p^2 - 1)! \cdots (p^p - 1)!}Are integers. Notice now that: \frac{b}{a} = \frac{p^2 \cdot p^3 \cdots p^p}{(p-1)!} \Rightarrow b = p^{2 + \cdots + p} \cdot \frac{a}{(p-1)!}So as p and (p-1)! are coprime we have that \frac{a}{(p-1)!} is an integer. Then so is: \frac{p \cdot a}{p!}As is said in the problem.