# Determining solutions to congruence equations

The first exercise states:

Suppose that $a$ and $b$ are integers, such that $a \equiv 12 \pmod{5}$ and $b \equiv 3 \pmod{5}$.

Given $c \equiv a \cdot b \pmod{5}$, find $c$ in $\mathbb{Z}_5$.

The second exercise states:

Suppose that $a$ and $b$ are integers, such that $a \equiv 11 \pmod{19}$ and $b \equiv 3 \pmod{19}$.

Given $c \equiv 8 \cdot b \pmod{19}$, find c in $\mathbb{Z}_{19}$.

For anyone that decides to help, could you please be as elementary as possible. I'm in my second week of Discrete Math and it's very easy for me to get lost. Thank you.

• Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Oct 12 '17 at 10:37

## 1 Answer

The value of $c \equiv a \cdot b \pmod{5}$ is the remainder when the product $ab$ is divided by $5$.

You can multiply, then reduce modulo $n$. Thus, \begin{align*} c & \equiv 12 \cdot 3 \pmod{5}\\ & \equiv 36 \pmod{5}\\ & \equiv 1 \pmod{5} \end{align*}
Also, you could use the property that if $a \equiv \alpha \pmod{n}$ and $b \equiv \beta \pmod{n}$, then $ab \equiv \alpha\beta \pmod{n}$ to simplify the calculations. Here, $12 \equiv 2 \pmod{5}$, so \begin{align*} c & \equiv 12 \cdot 3 \pmod{5}\\ & \equiv 2 \cdot 3 \pmod{5}\\ & \equiv 6 \pmod{5}\\ & \equiv 1 \pmod{5} \end{align*}

I will leave the second exercise to you.