Multiple-choice: sum of primes below $1000$ I sat an exam 2 months ago and the question paper contains the problem:


  Given that there are $168$ primes below $1000$. Then the sum of all primes
  below 1000 is 
  
  (a) $11555$ (b) $76127$ (c) $57298$ (d) $81722$


My attempt to solve it: We know that below $1000$ there are $167$ odd primes and 1 even prime (2), so the sum has to be odd, leaving only the first two numbers. Then I tried to use the formula "Every prime can be written in of the form $6n-1$,$6n+1$ except $2$ and $3$.", but I got stuck at that.
The sum of the first 168 positive integers is $\frac{168^2+168}{2}=14196$, which is greater than answer (a). The sum of the first 168 primes must be even greater than that.
The methods provided previously are very correct. However, I would like to share my approach as I believe this is even simpler. It's less calculative, and more logical.
Sum is odd ⇒ answer is either (a) or (b)
Taking the number in (a) 11555 we divide it by 168 [to get the average]
11555/168≈69
69&lt;168/2
An average of 168 distinct numbers can be less than 168/2(=84) iff at-least some of them are negative.
Which is clearly not possible since we are dealing with whole numbers.
⇒ only possible answer remain (b).