Converting Positive Integers - SS1 Mathematics Lesson Note

Given a positive integer $n$ to be converted to its equivalent $R$ in $modm$, thus,

$\frac{n}{m}=qremainder\mathbf{R}$,

Thus, $n=\mathbf{R}(modm)$

$n=q.m+R$, where $q$ is any positive whole number whose product with $m$ is as close as possible to $n$ without exceeding n. This is a statement to test if the congruence between $Randn$ is $\mathbf{true}$.

Example: Convert 51 to its equivalent in $mod10$

Solution

$n=51andm=10$

$$\frac{n}{m}=\frac{51}{10}=5remainder\mathbf{1}$$

Thus, $51\equiv \mathbf{1}(mod10)$ meaning $51$ is equivalent to $\mathbf{1}$ in $mod10$.

Checking with $n=q.m+R$

$51=5.10+1$

$51=50+1$

$51=51$, which is true, so we are correct.

Example: Find $x$ in the equation, $3\equiv x(mod4)$ (In essence, find the equivalent of ordinary $3$in$mod4$)

Solution

$n=3$ and $m=4$

$$\frac{3}{4}=0remainder\mathbf{3}$$

Thus, $3\equiv \mathbf{3}(mod4)$ and $x=\mathbf{3}$