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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 ?

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$6n=6(4q+2)=24q+12=4(6q+3)$ is cleanly divisible by $4$ and then some. – anon Jul 17 '12 at 2:00
up vote 2 down vote accepted

"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 $4$.

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Interesting ,could you tell me what would the remainder be in case the equation was not divisible by 4 and it was $6n=24k+13$ ? Would it have been $6n= 24k + 13 => 6n=24k+12+1$ with remainder 1 ? – Rajeshwar Jul 17 '12 at 2:54
If $n = 24k+13 \implies n = 4(6k+3) + 1$. Hence, in this case $n$ divided by $4$ gives $1$ as remainder. – Quixotic Jul 17 '12 at 16:15

Hint $\rm\ mod\ 4\!:\ n\equiv 2\:\Rightarrow\: 6n\equiv 6\cdot 2\equiv 0$

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