Question

If $a\equiv b\left(modn\right)$and $k|n$, is it true that role="math" localid="1646223432316" $a\equiv b\left(modk\right)$? Justify your answer.

Short answer

It is proved that $a\equiv b\left(modk\right)$.

Step-by-step solution

Definition of Congruence

Consider a, b, n as an integer with $n>0$. If ndivides $a-b$, then a iscongruent tob modulo n (expressed as $a\equiv b\left(modn\right)$).

Check whether a≡bmodk  is true or not

Consider that $a\equiv b\left(modn\right)$and $k|n$.

This signifies that there is an integer p in which the following is true because $k|n$:

$n=kp$

Moreover, $a\equiv b\left(modn\right)$.

This follows that there is an integer p in which the following holds:

$a-b=nq$

Put $n=kp$ in the above equation as follows:

$\begin{array}{rcl}a-b& =& kpq\\ & =& k\left(pq\right)\end{array}$

It further indicates that $a\equiv b\left(modk\right)$.

Thus, it is proved that $a\equiv b\left(modk\right)$.