Numbers which are sums of all smaller primes I have to find all numbers which are the sum of all smaller primes.
I think, that only number which is sum of all smaller primes is $5$ (all smaller primes than $5$ are only $2$ and $3$, and $2+3=5$).
If I will try to find such number by adding consecutive primes I will soon get too big numbers:
$2+3+5=10$, but $7$ is a prime smaller than $10$, so $10$ isn't a solution.
Similarly $2+3+5+7=17$, but there are $11$ and $13$ which are primes smaller than $17$, and so on, so it's looking as sum of consecutive primes is getting bigger too fast. On the other hand, I'm aware of the fact that there are "holes" without primes in any size (there are n consecutive natural numbers somewhere without primes, for any n), but I think it will not change anything.
So I conjecture, that $5$ is only solution, but unfortunately I have no idea how to prove it mathematically.
 A: Let $\Psi(n)$ denote the sum of the primes less than $n$.  We contend that $n>5$ implies $\Psi(n)>n$.  That is clearly stronger than the desired result.
It is easy to confirm that this is true for the first few $n$.  We then proceed by induction.
Suppose $N$ is the least counterexample.  Then, in particular, the claim must be true for $\Big \lfloor \frac N2\Big \rfloor$ (we may assume that we have checked far enough so that $\frac N2>6$).  But then $$\Psi\left(\Big \lfloor \frac N2\Big \rfloor\right)>\frac N2$$
By Bertrand there is a prime $p$ between $\frac N2$ and $N$ so that we have $$\Psi(N)≥\Psi\left(\Big \lfloor \frac N2\Big \rfloor\right)+p>N$$ and we are done.
A: Use the notation $p_k$ to be the $k$th prime.
By Bertrand's Postulate then for any $p_k$ then $p_k < p_{k+1} < 2p_k$.
Suppose $(p_1 + p_2 + .... + p_k) > p_{k+1}$ then for any number $M_k: p_k < M \le p_k+1$ then $M_k$ fails to be such a number.
And as $(p_1 + p_2 + .... + p_k) + p_{k+1} > 2p_{k+1} > p_{k+2}$ then for any number $M_{k+1}: p_{k+1}< M_{k+1}\le p_{k+2}$ then $M_{k+1}$ fails to be such a number.
And that's it.  We can prove our result with induction.
$P(n)$ will be the statement:  if $5 < M \le p_n$ then $M$ fails to be such a number.
We well prove $P(n)$ is true for all $n\ge 4; p_4=7$, and thus all $M > 5$ fail.
Base case $n=4$ then for $M: 5< M \le 7$ then $M < 2+3+5=10$.
Induction step: Assume $P(k)$.  
So for any $M:5<M \le p_k$ we know $M$ fails.
So for and $5\le p_{k-1}< M \le p_k$ we know $M$ fails so it fails for $p_k$.
$p_1+p_2 + .... + ,p_{k-1}<p_k$ and $(p_1+p_2 + .... + p_{k-1})+p_k > 2p_k > p_{k+1}$.  So it fails for all $p_k < N\le p_{k+1}$.
So it fails for all $M: 5 < M \le p_{k+1}$.
So $P(k+1)$
.....
Thus $P(n)$ for all $n\ge 4$ and the we have proven by induction that for all $M > 5$ this fails.
A: Using @lulu hint:
It's easy to check that there are no more such numbers up to 30. Now let's take $n\ge25$. We will check that then any of the numbers $\lceil(1+\frac{1}{5})n\rceil, \lceil(1+\frac{1}{5})n\rceil+1, \lceil(1+\frac{1}{5})n\rceil+2, \lceil(1+\frac{1}{5})n\rceil+3, \lceil(1+\frac{1}{5})n\rceil+4$ are not satisfying conditions of the exercise. 
From Jitsuro Nagura result (1952), there is a prime, name it $p_{1}$, satisfying $n<p_{1}<(1+\frac{1}{5})n$. Now, the Bertrand's postulate says that there is a prime, name it $p_{2}$, with $\frac{n+2}{2}<p_{2}<n$. But $p_{1}+p_{2}>n+\frac{n+2}{2}=\frac{3n}{2}+1$ and for $n\ge25$ $$\frac{3n}{2}+1>(1+\frac{1}{5})n+8>\lceil(1+\frac{1}{5})n\rceil+4,$$ so only number which is the sum of all primes smaller than this number is 5.
