Question

Given the congruence 12x (congruent to) 8(mod 7).

a. Determine if there is an inverse of 12 modulo 7. If there is an inverse, find it.
b. Determine if the congruence 12x (congruent to) 8(mod 7) has a solution for x and if so, find a solution

Answer 1

Well, \(8\pmod 7\) is no different then \(1\pmod 7\) right?

Answer 2

\(8=1+7k,\,k=1\)

Answer 3

I believe for this you would use extended Euclidean algorithm.

Answer 4

\[
\color{red}{12}x=1\pmod {\color{orange}{7}}\\
\begin{array}{cccc}
\color{orange}{7} &=& \color{red}{12}&\times& 0 &+&\color{blue}{7}\\
\color{red}{12} &=& \color{blue}{7}&\times& 1&+& \color{green}{5}\\
\color{blue}{7} &=& \color{green}{5} &\times& 1 &+& 2\\
5 &=& 2&\times& 2&+&1
\end{array}
\]
The purpose is that this gets us the following equations\[
\begin{array}{ccccc}
1&=&5-2\times 2\\
2&=&7-5\times 1 &=& 7-5\\
5&=&12-7\times 1&=&12-7\\
7&=&7-12\times 0 &=& 7
\end{array}
\]We can substitute them: \[
\begin{split}
1&=5-2\times 2 \\
&=5-(7-5)\times 2 &=3\times 5-2\times 7 \\
&=3\times (12-7) - 2\times 7 &= 3\times 12-5\times 7
\end{split}
\]This gives us: \[
3 \times 12 = 5\times 7+1 \implies 3\times 12 =1 \pmod 7
\]So we end up with \(x=3\).

Answer 5

How do you determine if there is an inverse of 12 modulo 7?

Answer 6

how is 8(mod 7) is same as 1 (mod 7)?

Answer 7

\[
8 = 1 + 7k
\]Where \(k=1\).

Answer 8

I am not sure if I understand where did you get 8 = 1 +7k from...

Answer 9

\(8\equiv 1\pmod{7}\) since you're left with a remainder of 1 when you divide 8 by 7.

Furthemore, when you say something like \(a\equiv b\pmod{n}\), it's equivalent to saying that \(n\mid a-b \implies a-b = nk\implies a = b+nk\) (for some \(k\in\mathbb{Z}^+\))

Hence, when we say that \(8\equiv 1\pmod{7}\), we can express 8 as 1 plus a multiple of seven, i.e. \(8=1+7k\); in this particular instance, the only k that satisfies this equation is \(k=1\).

Does this clarify things with regards to that part? :-)