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I am having problems solving this question :

When n is divide by 4 the remainder is 2 what will the remainder be when 6n is divided by 4 ? Ans=$0$

Here is what I have got so far

$\frac{n}{4} => Remainder ~ 2$ so we get $n=4q+2$

$\frac{6n}{4} => Remainder ~ ?$ so we get $6n=4p+r_{emainder}$

How do we solve for remainder here ?

  • 4
    $6n=6(4q+2)=24q+12=4(6q+3)$ is cleanly divisible by $4$ and then some.2012-07-17

2 Answers 2

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"When n is divide by $4$ the remainder is $2$": $ n \equiv 2 \pmod 4 \text{ or } n = 4k+2, k \in \mathbb{Z}$

For $6n$ we have, $ 6n = 24k+12 = 4(6k+3) +\color{red}{0} = 4m+\color{red}{0}, m = (6k+3) \text{ (an integer) } $

Hence, $0$ will the remainder be when $6n$ is divided by $4$.

  • 0
    If $n = 24k+13 \implies n = 4(6k+3) + 1$. Hence, in this case $n$ divided by $4$ gives $1$ as remainder.2012-07-17
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Hint $\rm\ mod\ 4\!:\ n\equiv 2\:\Rightarrow\: 6n\equiv 6\cdot 2\equiv 0$