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.
 A: Primes except for $2$ are all odd, and you have $168$ distinct primes, so their sum must be at least $2 + \sum_{k=2}^{168} (2k-1) = 2 + (168^2-1) > 160^2 = 25600 > 11555$. So option (a) is out and only (b) remains.
A: you just have to decide between $11555$ and $76127$.
Notice that the first implies the average prime under $1000$ is $11555/168<69$. Which is clearly false.
A: We have to decide among $\text{(A)}$ and $\text{(B)}$. Note that the $26$th prime is $101$. This implies that if $p_{n}$ denotes the $n$ th prime, then $$\sum_{n=1}^{168}p_{n} = \sum_{n=1}^{25}p_{n}+\sum_{n=26}^{168} p_{n} > \sum_{n=26}^{168} 101 =101 \times 143=14443 >\text{(A)}=11555$$
The answer is thus $\text{(B)}$, $76127$. The answer can be confirmed through direct calculation or can be verfied here. 
A: There are $168$ primes with the first one equal to $2$ the rest $\ge 2k-1$ for $k=2,3,4,...,168$.   So their sum is at least $168^2+1=28 225$.
A: 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.
A: I just wanted to carry forward your observation about "Every prime can be written in of the form $(6n-1),(6n+1)$ except $2$ & $3$".
We can quickly get a minimum sum out of this. Assume that the $166$ primes not $2$ or $3$ are the smallest such numbers obeying the above; then $83$ are $6k{-}1$, $83$ are $6k{+}1$ and the minimum bound total is $83$ terms of $12k$, which is $12\cdot 84 \cdot 83 /2 = 504\cdot 83  = 41832$ - and we can decoratively add the $2$ and $3$ to get $41837$. This is more than big enough to eliminate option (a) as required.
A: Your analysis that the answer must be $(a)$ or $(b)$ is convincing.  Considering that $(a)$ and $(b)$ are much different, just about any simplistic method to approximate the sum of all primes should tell which is the right answer.
For example, the sum of all numbers less than $1000$ is about $500,000$. So, $\cfrac{168}{1000} \times 500,000$ or $84,000$ should be in the right ballpark.  $76127$ is the right answer, by this reasoning.
A: 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<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).
