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
-1
votes
2 answers
solving congruence class equation
Keeping $p = 7$ is the following true?
If for any congruence class $[a]$ that is not $[0]$, there is a unique
class $[b]$ such that $[a] \cdot [b] = [1]$.
Would this be true? (assuming you keep $p = 7$)
M.Jones
- 357
-1
votes
2 answers
About primitive root and quadratic residue.
If p is an odd number and g is a primitive root (mod p), how can I prove that g is not a quadratic residue mod p?
A.Zhou
- 1
-1
votes
2 answers
How would I go about solving the following simultaneous congruence:
How do I solve the following simultaneous congruence:
$25x \equiv 18 (\mod 48)$
$x \equiv 11 (\mod 35)$.
dd123
- 45
-1
votes
1 answer
find the solution of $7x \equiv 29\pmod{31}$
Find the solution of $7x \equiv 29\pmod{31}$.
I have this solution but not understand it:
Since $9$ is the inverse of $7 \bmod 31$, multiply both sides by $9$.
Then $9\cdot 7 x \equiv 9\cdot 29 \pmod{31}$. And therefore
\begin{eqnarray*}
x &\equiv&…
user155971
- 1,515
-1
votes
1 answer
Prove that if $9\mid(x^3 + y^3 + z^3)$ then $3\mid xyz$ for integers $x, y, z$
how would I prove that if 9 divides $x^3+y^3+z^3$ then $3$ divides $xyz$?
I've thought about it and don't really know where to start. So far, I know that $x^3+y^3+z^3$ is congruent to $0 (\mod 9)$.
I don't know where to go from here.
DayDreamerz
- 37
-4
votes
2 answers
Primitive root modulo $p$.
Let $a$ be an integer and $p$ an odd prime number. Suppose $\bar{a}$ has order $h>1$ in $\mathbb{Z}_p^*$, how can we show that $$a^{h-1} + a^{h-2} + ... + a + 1\equiv 0\mod p?$$
user426865
- 15
-6
votes
1 answer
Discrete math question: congruence
I am trying to solve the following question:
Determine $n$ between $0$ and $19$ such that $(2311)(3912) = n \mod 20$.
Where I am now:
(2311) = 11(mod 20)
(3912) = 12(mod 20)
Multiply the 11X12 to get 132;
So:
(2311)(3912) = 132(mod 20)
My question…
user401806
- 21