The tricky sum

The sum of the first $n$ whole numbers is $\frac{n(n-1)}{2}$ . Replacing $n$ with 2014 gives the answer 2027091 so the last three digits are 091. Finally, note that 0 is a whole number.

Point to note: Whole numbers are non-negative integers, meaning they start from $0, 1, 2...$ and not from $1, 2, 3...$ .

Hence, the sum of the first 2014 whole numbers is $0 + 1 + 2 + ... + 2011 + 2012 + 2013 = \frac{2013 \times 2014}{2} = 2027091$

The last 3 digits of 2027091 is $\boxed {091}$ .

The "tricky" part of this is that the question asks for whole numbers, which includes $0$ . What we're searching for is actually the sum of the first $2013$ natural numbers, which is given by:

$\dfrac{2013 \times (2013+1)}{2} = 1007 \times 2013 = 2027091 \Longrightarrow \boxed{091}$