For questions about congruences in modular arithmetic, concerning for example the chinese remainder theorem, Fermat's little theorem or Euler's totient theorem.
Questions tagged [congruences]
1724 questions
9
votes
3 answers
Prove that a number is divisible by 3 iff the sum of its digits is divisible by 3
I'm trying to teach myself some number theory. In my textbook, this proof is given:
Example (2.3.1) Show that an integer is divisible by 3 if and only if the sum of its digits is a multiple of 3.
Let $n=a_0a_1\ldots a_k$ be the decimal…
John L.
- 273
7
votes
3 answers
Number of solutions for $x^2 \equiv x \pmod m$
What is the number of solutions of $x^2 \equiv x \pmod m$ for any positive integer $m$?
Yogesh Ghaturle
- 305
5
votes
4 answers
how to prove $m^{\phi(n)}+n^{\phi(m)}\equiv 1 \pmod{mn}$ where m and n are relatively prime?
I do not know how to prove this question: $m^{\phi(n)}+n^{\phi(m)}\equiv 1 \pmod {mn}$ where $m$ and $n$ are relatively prime.
Can anyone help?
Sayantan Koley
- 638
4
votes
5 answers
$x^n =2($mod $13)$
Consider the congruence $x^n =2($mod $13)$. Then for which n does it have a solution?
n= 5
n=6
n=7
n=8
Can we take any help from this fact $x^{12 }=1 ($ mod $13),\forall x $?
Please give some hints.
Sankha
- 1,405
4
votes
3 answers
How to solve $x=2^{-18}$ (mod 143)
I have to solve the following equation: $x=2^{-18} \mod 143$.
The problem is that I can't use Fermat's little theorem as $\varphi(143)=120$ which doesn't help at all. The other method I know is to find the inverse of $2^{18} \mod 143$ using Euclid's…
KeykoYume
- 492
4
votes
1 answer
$\frac{(p-1)!}{1}-\frac{(p-1)!}{2}+\frac{(p-1)!}{3}-\cdots-\frac{(p-1)!}{p-1} \equiv \frac{2-2^p}{p} \pmod{p}$
$p$ is an odd prime. How to prove the following congruence?
$$\frac{(p-1)!}{1}-\frac{(p-1)!}{2}+\frac{(p-1)!}{3}-\cdots-\frac{(p-1)!}{p-1} \equiv \frac{2-2^p}{p} \pmod{p}$$
I have created a polynomial that the left side of this congruence is one…
user120269
3
votes
2 answers
Chinese remainder problem
$\begin{cases}
x \equiv 39 \pmod{189}\\
x \equiv 25 \pmod{539}\\
x \equiv 39 \pmod{1089}\end{cases}$
but two moduli are not pairwise prime $(189, 1089)=3$
What do we do to solve it then? Should we try to solve $x\equiv 39 \pmod{189}$ and $x\equiv…
user124471
- 620
3
votes
4 answers
Prove that for all integers $n \ge 0$, $5^n \equiv 1+4n\pmod {16}$
That $5^n$ really threw me off and I'm not really sure how to start this problem.
What can I do in a congruence problem with the exponent $n$?
Haxify
- 435
- 1
- 3
- 8
3
votes
4 answers
$74x \equiv 1 \bmod 53$
I've figured out that $x=-5$ for this to work but I don't know how to state the general solution? (all cases)
Anyone know? D:
user122661
- 983
3
votes
0 answers
Find a solution to the congruence $x^3 \equiv$ $4 $ mod $ 5$
My question
So there are three parts;
(a) Find a solution to the congruence $x^3 \equiv$ $4$ mod $5$ by trying all possible $x$ values.
(b) Find a solution to the congruence $x^3 \equiv$ $4$ mod $11$ by trying all possible $x$ values.
(c) Find a…
Rubicon
- 616
3
votes
3 answers
Can a be written k + b for all integers a and b?
Instinctively I think yes, since $a \equiv b$ (modulo 1) for all $a$ and $b$.
Farman H
- 103
3
votes
2 answers
Solving quadratic or higher degree congruence with very large modulus.
Is there any general way to solve a polynomial congruence with a very large modulus?
An example could be
$$
x^2-377x+1\equiv 0 \pmod {8683317618811886495518194401279999999 }
$$
or
$$
x^2-29478x\equiv 13^{67} \pmod {17^{2001}}
$$
YoTengoUnLCD
- 13,384
3
votes
2 answers
How to solve polynomial Congruence
$21x^5 + 10x^3 + 14x \equiv 0 \pmod{3}$.
I want to use flt and crt to solve this but I am not sure where to start. Do you have any tips?
Joe
- 91
2
votes
2 answers
Congruence-What is remainder when $a^2+2b$ is divided 3?
Suppose $a\equiv 7 \pmod 9 $, $b\equiv 1\pmod 6$. What is remainder when $a^2+2b$ is divided by 3 ?----- So, since $a=9m+7$, $b=6n+1$, $m,n\in \mathbb{Z}$, $a^2+2b=(9m+7)^2+2(6n+1)$,=$81m^2+12n+177$=$3(27m^2+4n+59)$....I am not sure what I am…
Wes
- 497
2
votes
2 answers
Properties of Congruences
For context:
The question/answer is given as follows:
Show by induction that if n is a positive integer, then $4^n ≡ 1 + 3n (\text{mod } 9)$.
For the base case, $4 ≡ 1+3 (\text{mod } 9)$.
For the induction hypothesis, assume that $4n ≡ 1+3n…
Alex Luecke
- 23