A whole number $n$ when divided by $4$ gives $3$ as a remainder. What will be the remainder when $2n$ is divided by $4$?

According to the problem, the number $n$ can be expressed in the form:
$$n=4k+3$$

where $k$ is a whole number quotient.

Now, multiply the expression by $2$ to find $2n$:

$$2n=2(4k+3)$$

$$2n=8k+6$$

We can rewrite $8k+6$ to explicitly show the multiples of $4$:

$$2n=8k+4+2$$

$$2n=4(2k+1)+2$$

When $4(2k+1)+2$ is divided by $4$, the term $4(2k+1)$ is perfectly divisible by $4$, leaving a remainder of $2$.

Alternative Method (Modular Arithmetic):

$$n\equiv 3(\mathrm{mod}4)$$

$$2n\equiv 2\times 3(\mathrm{mod}4)$$

$$2n\equiv 6\equiv 2(\mathrm{mod}4)$$

Thus, the remainder is $2$.