Linked Questions

Prove that sum of elements in reduced residue system modulo $n \in N$ is divisible by $n$. I feel like problem just comes down to pairing elements of RRS in way that they are congruent, but can't ...

Given $m > 0$, fix a reduced residue system (RRS) $r_{1}, r_2,\dotsc , r_{\varphi(m)} $ mod $ m$. Let $x$ denote the sum $r_1 + r_2 + \dotsb + r_{\varphi(m)}$. What is $x$ mod $m$? The problem is ...

I want to prove why $\phi(n)$ is even for $n>3$. So far I am attempting to split this into 2 cases. Case 1: $n$ is a power of $2$. Hence $n=2^k$. So $\phi(n)=2^k-2^{k-1}$. Clearly that will ...

This should be an easy exercise: Given a finite odd abelian group $G$, prove that $\prod_{g\in G}g=e$. Indeed, using Lagrange's theorem this is trivial: There is no element of order 2 (since the order ...

Is there a way to prove that the sum of the arithmetic progression $a_1, a_2, \dots, a_n$ can be calculated by $\displaystyle s_n = \frac{n}{2}(a_1+a_n)$?

I can't figure this problem out. Let $b_1, b_2, \cdots, b_{\phi(m)}$ be the integers between $1$ and $m$ that are relatively prime to $m$. $B$ is the product of these integers. Prove that either $B$ ...

For example, the sum of 1, 2, and 3 (three consecutives -- an odd number) is 6, which is a multiple of 3.

I understand that for a number $a$ to have a multiplicative inverse in mod $n$, it must be coprime to $n$; therefore, all numbers that do not have a multiplicative inverse mod $n$ must share a factor ...

Question 12(iii) Could anyone explain this part of the question to me. What i tried co-prime means that the two integers a and b are said to be relatively prime, mutually prime, or coprime (also ...

In base $10$, all prime numbers (a part $2$ and $5$) end with $1,3,7$ or $9$, i.e. with four different symbols. Is there a base in which all prime numbers end with $5$ different symbols (or also with ...