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(B) Given that $n$ is a whole number such that $n = 4k + 3$ for some integer $k ge 0$.We need to find the remainder when $2n$ is divided by $4$.Substi

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$2$

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(B) Given that $n$ is a whole number such that $n = 4k + 3$ for some integer $k ge 0$.We need to find the remainder when $2n$ is divided by $4$.Substitute the expression for $n$ into $2n$:$2n = 2(4k + 3) = 8k + 6$.Now,divide $8k + 6$ by $4$:$8k + 6 = 4(2k) + 4 + 2 = 4(2k + 1) + 2$.Since $4(2k + 1)$ is perfectly divisible by $4$,the remainder is $2$.Illustration: Let $n = 7$ (which leaves a remainder of $3$ when divided by $4$).Then $2n = 14$.Dividing $14$ by $4$ gives $14 = 4 	imes 3 + 2$,so the remainder is $2$.

$n$ is a whole number which when divided by $4$ gives the remainder $3$. The remainder when $2n$ is divided by $4$ is

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