Of course $f(p,n)=v_p(n!)=\sum_{i=0}^d a_i(1+p+\cdots+p^{i-1})=\sum_{i=0}^d a_i\frac{p^i-1}{p-1}$ where $n=a_d....a_1a_0$ is the representation of $n$ in base $p$. We now use the following lemma: for each $m$ there exists $N$ such that $m|\frac{p^N-1}{p-1}$. Proof: if $q|m$ is prime and $q\nmid p-1$, take $\phi(q^{v_q(m)})|N$, so that $p^N\equiv 1\pmod{q^{v_q(m)}}$. If $q|p-1$ take $q^{v_q(m)}|N$, so that $v_q(\frac{p^N-1}{p-1})=v_q(N)\geq v_q(m)$ by LTE (if $q=2$ we can do analogously) Now by our lemma take $n=\sum_{i=0}^{c-1} p^{iN+1}$. In this way $f(p,n)=\sum_{i=0}^{c-1} 1+p\frac{p^{iN}-1}{p-1}\equiv \sum_{i=0}^{c-1} 1+p*0=c\pmod{m}$, so we are done

Can't we use the infinite PHP? Assume there exist only finitely many such $n$ for some choice of $p,m,c,$ i.e. let $f(p,n)\neq c\pmod m$ for all $n > n_0.$ Then \begin{align*} f(p, n_0 + 1) &= f(p, n_0) + v_p(n_0 + 1) = c + v_p(n_0 + 1) \pmod m \\\sum_{i = 1}^{k} v_p(n_0 + i)&\neq 0\pmod m \end{align*}for all $k\in\mathbb{N}.$ Can we use this in some way?

You seem to assume that for $n=n_0$, you actually have what you want. In other words, you assume that already one such $n$ exists. Indeed, once you show that one such $n$ exists, it is much easier (though not yet trivial!) to find infinitely many such $n$. However, all the constructions of one such $n$ that I know, already immediately construct infinitely many such $n$.

Hello guys

Jasurbek wrote: Xullas bir tinga qimmat yechim ekan tushunarli qilib yozilsa bo'midimi??? Welcome to this forum. Please ask your question in English so that we have a chance to understand it.

cadaeibf wrote: Of course $f(p,n)=v_p(n!)=\sum_{i=0}^d a_i(1+p+\cdots+p^{i-1})=\sum_{i=0}^d a_i\frac{p^i-1}{p-1}$ where $n=a_d....a_1a_0$ is the representation of $n$ in base $p$. We now use the following lemma: for each $m$ there exists $N$ such that $m|\frac{p^N-1}{p-1}$. Proof: if $q|m$ is prime and $q\nmid p-1$, take $\phi(q^{v_q(m)})|N$, so that $p^N\equiv 1\pmod{q^{v_q(m)}}$. If $q|p-1$ take $q^{v_q(m)}|N$, so that $v_q(\frac{p^N-1}{p-1})=v_q(N)\geq v_q(m)$ by LTE (if $q=2$ we can do analogously) Now by our lemma take $n=\sum_{i=0}^{c-1} p^{iN+1}$. In this way $f(p,n)=\sum_{i=0}^{c-1} 1+p\frac{p^{iN}-1}{p-1}\equiv \sum_{i=0}^{c-1} 1+p*0=c\pmod{m}$, so we are done But what if $p|m$?