Question: Find the numbers of integers such that both 5n and n/3 are 4-digit integers.
Minimum 4-digit integer = 1000
Maximum 4-digit integer = 9999
$$Therefore,\ \frac{1000}{3}\le\frac{3n}{3}\le\frac{9999}{3}$$
$$333.33\le n\le3333\ \ \ \ ---eqn\ \left(1\right)$$
$$1000\le\frac{n}{3}\le9999$$
$$1000\cdot3\le\frac{n}{3}\cdot3\le9999\cdot3$$
$$3000\le n\le29,997\ ---eqn\left(2\right)$$
Combining both equation together,
Since 333.33 and 29,997 are not 4-digit integers, we will have $$3000\le n\le3333$$
For 'n/3' to be an integer, it must be divisible by 3. So, we use
$$=\frac{upper\ range-lower\ range}{3}+1$$
$$=\frac{3333-3000}{3}+1$$
$$=\frac{333}{3}+1$$

Therefore, the correct answer is option B.