### Bellala's blog

By Bellala, history, 5 weeks ago,

615D - Multipliers

The way to solve this problem is easy. Just let $n=p_1^{c_1}p_2^{c_2}\cdots p_k^{c_k}$ ($p_i$ are pairwise different) then the answer will be

$ans=\prod_{i=1}^k \left(p_i^{1+2+\cdots +c_i}\right) ^ { (c_1+1)\cdots (c_{i-1}+1)(c_{i+1}+1)\cdots (c_k+1) } \mod p \\= \prod_{i=1}^k \left(p_i^{c_i(c_i+1)/2}\right) ^ { (c_1+1)(c_2+1)\cdots (c_{i-1}+1)(c_{i+1}+1)\cdots (c_k+1) \mod p-1} \mod p$

There're more than one way to get ans (a simple way is to use prefix product). But today I read a solution which I could not understand:

solution

My only question: why we can keep the factor 2 by mod 2*(mod-1)? Thanks in advance

• +3

 » 5 weeks ago, # |   0 It's actually a quite fancy trick IMO. I tried to explain as detailed as I can:Let $x$ denote $\prod{(c_j + 1)}$. We know that $A = c_i \cdot x$ is always even since both $c_i$ and $c_i + 1$ are present in that product (exactly one is even). So, $\frac{A}{2}$ is an integer. (We already knew it because it must be even somehow but I wanted to clarify this part also.)Let's denote $p-1$ by $m$ for convenience. We want to find the remainder of $\frac{A}{2}$ divided by $m$, that is, the non-negative integer $r < m$ such that $\frac{A}{2} = km + r$ where $k$ is an integer.We can multiply both sides by $2$ to obtain $A = k \cdot 2m + 2r$. So, if we can obtain a number congruent to $A \bmod 2m$, it is going to be even (as both $A$ and $2m$ are even).Notice that $A' = c_i \cdot (x \ \% \ 2m)$ is congruent to $A \bmod 2m$, so, $A'$ can also be written as $A' = k' \cdot 2m + 2r$, which indicates that $\frac{A'}{2} = k'm + r$. As a result, it is safe to divide $A'$ by $2$ and then take $\frac{A'}{2}\bmod m$.