Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$\begin{align} (x-1) \;\text{mod}\; 11 &= 3x\; \text{mod}\; 11\\ 11&\lvert(3x-(x-1)) \\ 11&\lvert2x+1\\ x &= 5?\\ \end{align} $$ $$ \begin{align} (a-b)\; \text{mod}\; 5 &= (a+b)\;\text{mod}\;5\\ 5&\lvert a+b-a+b\\ 5&\lvert2b\\ b &= 5/2\\ a &= \text{any integer} \end{align} $$ I don't know how to solve this type of problem. Can you tell me what I have to do generally step-by-step?

share|cite|improve this question
I edited your question, I hope that I didn't introduce any mistakes. – Thomas Dec 7 '12 at 15:06

For the first one you have correctly that $11 \lvert 2x + 1$. What does this means? This means that there is some integer $n$ such that $$ 11n = 2x + 1. $$ This is equivalent to $$ x = \frac{11n - 1}{2}. $$ Note here that for $x$ to be an interger we need $11n - 1$ to be even. This happens exactly when $n$ is odd. And when $n = 2m + 1$ is odd, then $$ x = \frac{22m + 11 - 1}{2} = 11m + 5. $$ So the solution is all those integers $x$ of the form $11m + 5$ where $m$ is any integer.

For the second one, you do likewise starting with $$ 5\lvert 2b. $$ Again this means that there is some integer $n$ such that $$\begin{align} 5n &= 2b & \Rightarrow\\ b &= \frac{5}{2}n. \end{align} $$

Now this only gives you an integer if $n = 2m$ is even, and in that case you get $$ b = 5m. $$ So the solution is all those pairs of integers $(a,b)$ such that $b = 5m$ for some integer $m$. (It is therefore correct when you note that $a$ can be any integer).

share|cite|improve this answer

There are several ways; for example look at Diophantine divisors, but now I will write it ;

$ax\equiv b\;(mod\;m)\Longleftrightarrow (a,m)=d|b$ and its answers are every number in congreuent classes by modulo m like $(\frac{a}{d})^{*}(\frac{b}{d})+k\frac{m}{d}$ where $0\leq k\leq d-1$ and $(\frac{a}{d})^{*}$ in Möbius inversion of $\frac{a}{d}$ in mod $\frac{m}{d}$

When $(a,m)=d|b$, we can say $(\frac{a}{d},\frac{m}{d})=1$ and from Euler theorem that $a^{\phi(m)}\equiv 1\;(mod\;m)\Longrightarrow a^{\phi(\frac{m}{d})-1}a\equiv 1\;(mod\;\frac{m}{d})$ so $a^{\phi(\frac{m}{d} )-1}$ is one Möbius inversion of $\frac{a}{d}$ in mod $\frac{m}{d}$ . By finding Möbius inversion, and replacing in $(\frac{a}{d})^{*}(\frac{b}{d})+k\frac{m}{d}$, then every number in congreuent classes by modulo m that is found, are the answers you need.

For example let calculate your first question;

$x-1\equiv 3x\;(mod\;11)\Longleftrightarrow 2x\equiv -1\;(mod\;11)$

because $(2,11)=1|-1$ so it has answer. Here $d$ is $1$.

At first calculate Möbius inversion of $2$ in mod $11$, (note that $d=1$!).

$2^{*}=2^{\phi(11)-1}=2^{10-1}=2^{9}=512\equiv 6\;(mod\;11)$

so $2^{*}=6$ now because $d-1=0$ we only put $k=0$ then answers is $\{x\in\mathbb{Z}|[x]_{11}=[6(-1)+(0)11]_{11}\}=$ $\{x\in\mathbb{Z}|[x]_{11}=[-6]_{11}\}=$ $\{x\in\mathbb{Z}|[x]_{11}=[5]_{11}\}=$ $\{11n+5|n\in\mathbb{Z}\}$ .

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.