Residue of $n$ given residue of $4n$

Question

If we divide a number by 5 the remainder is 3 (or 2 it is not clear) , so the remainder of division of quarter of the number by 5 is equal ......

Choose : a ) 0 , b ) 1 , C ) 2 , D ) 4

Answer 1

Note that $4\cdot 4 \equiv 1 \pmod 5$ so "dividing by $4$" is the same as multiplying by $4$ (modulo $5$).

Now let $4n$ be the number you speak about.

$$4n \equiv 3 \pmod{5} \qquad (\text{resp. }4n \equiv 2 \pmod{5})$$

Then

$$n \equiv 3\cdot 4^{-1} \pmod{5} \qquad (\text{resp. }n \equiv 2\cdot 4^{-1} \pmod{5})$$

i.e.

$$n \equiv 3\cdot 4 \equiv 2\pmod{5} \qquad (\text{resp. }n \equiv 2\cdot 4 \equiv 3 \pmod{5})$$

Alternatively, you can work the problem "backwards" from the options given:

$$4\cdot 0 \equiv 0 \pmod 5 \qquad 4\cdot 1\equiv 4\pmod 5$$ $$\color{red}{4\cdot 2 \equiv 3 \pmod 5}\qquad 4\cdot 4 \equiv 1\pmod 5$$.

Since the original number has residue $3$ modulo $5$, it follows that a fourth of that number has residue $2$.