$c \in \mathbb{N}$, so that $c \cdot 11 = 23 \mod 103$

Question

How can one find $c \in \mathbb{N}$, so that $c \cdot 11 = 23 \mod 103$?

I know that $a \cdot b \mod n = (a \mod n \cdot b \mod n) \mod n$.

Furthermore, Fermat's little theorem says that for all prime $p$ and if $p$ does not divide $a$

$a^{p-1} \equiv 1 \pmod p$

And I know that here all numbers $11, 23, \text { and } 103$ are prime numbers.

That still doesn't help me though...

$103$ is a prime. Also, $11$ is a prime, more specifically, it's relatively prime to $103$. Do you know how to compute multiplicative modular inverses? We have $c\equiv 23\times 11^{-1}\pmod{103}$ and $11^{-1}$ is computed by solving $11z\equiv 1\pmod{103}$ — Prasun Biswas, Nov 22 '18 at 10:52
My computations give me: $$11z\equiv 1\iff 7z\equiv 110z\equiv 10\iff 105z\equiv 2z\equiv 150\iff z\equiv 75\pmod{103}$$ which gives $c\equiv 77\pmod{103}$ — Prasun Biswas, Nov 22 '18 at 10:56

score 0 · Accepted Answer · answered Nov 22 '18 at 11:10

Fermat's little theorem provides a solution:

$$ c \cdot 11 \equiv 23 \bmod 103 \implies c \equiv c \cdot 11^{102} \equiv 23 \cdot 11^{101} \bmod 103 $$

Thus, $c = 23 \cdot 11^{101}$ works. It is not the smallest solution, but the question does not ask for that...

score 0 · Answer 2 · answered Nov 22 '18 at 19:50

0

By Gauss's algorithm $\!\bmod 103\!:\, \ c\equiv \dfrac{23}{11}\equiv \dfrac{9\cdot 23}{9\cdot 11}\equiv\dfrac{1}{-4}\equiv \dfrac{104}{-4}\equiv -26\equiv 77$

answered Nov 22 '18 at 19:50

Bill Dubuque

272,048

$c \in \mathbb{N}$, so that $c \cdot 11 = 23 \mod 103$

2 Answers2