find a solution of 9x = 24 (mod 21)

Question

I need help finding a solution of $9x\equiv {24}\pmod {21}$.

Here is what I tried, but it's wrong.

mod x is the positive value of x. mod $21 = 21.$

$9x\equiv {24}\pmod {21}$.

$9x = 24*21$

$x = 24*21/9 = 56$

Answer 1

$$9x=24\pmod{21}\iff 9x=24+21k=3+21m\;,\;k\,,\,m\in\Bbb Z\implies$$

$$3x=1+7k\implies x=5\pmod 7$$

since $\,3\cdot 5=15=1\pmod 7\iff 3^{-1}=5\pmod 7\;$

Answer 2

I'm not sure what you mean by "mod x is the positive value of x", but you should not multiply $24$ and $21$. What you want is a number $x$ such that $21$ will divide $9x - 24$. We write this as $21 \mid 9x - 24$. Note that $3$ divides both sides, so it must be that $21/3 \mid (9x - 24)/3$ or $7 \mid 3x - 8$. From there you can try and calculate an inverse to $3$ modulo $7$ or you can just do a little guess and check to find that $x = 5$ works.

Answer 3

Since $21=3\cdot 7$ and $(3,7)=1$, we have: $$9x\equiv 24\pmod{21}\Leftrightarrow 9x\equiv 24\pmod{3}\wedge9x\equiv 24\pmod{7}$$ Now, $9x\equiv 24\pmod{3}\Leftrightarrow 0x\equiv 0\pmod{3}$, which is always true and $$9x\equiv 24\pmod{7}\Leftrightarrow 2x\equiv 3\pmod{7}\Leftrightarrow 4\cdot2x\equiv 4\cdot 3\pmod{7}\Leftrightarrow x\equiv 5\pmod{7}$$

Answer 4

This essentially becomes a Diophantine equation:

$$9x\equiv{24}\pmod{21} \\ 9x-24=21y \\ 9x-21y=24 \\ 3x-7y=8$$

Solve the Diophantine equation: $3u-7v=1$ using the Euclidean algorithm:

$$3(2)+1=7\\ 3(-2)-7(-1)=1\\ u=-2;\,v=-1$$

Now multiply out by $8$ to get the original Diophantine equation:

$$3(8u)-7(8v)=1(8)$$

$$x=8u=-16$$

Therefore $x=-16$.