The sum of all integers from $1$ to $p$ is divisible by $p$ and all prime numbers before $p$. What can $p$ be? 
The summation of all integers from $1$ to $p$ is divisible by $p$ and all prime numbers before $p$. Find all possible solutions for $p$, with proof. Here $p$ is a prime number.   

This is a preparation problem for the upcoming BdMO. I can't find a way to start. Any hint will be helpful. 
 A: $$\sum_{k=1}^p k=\frac{p(p+1)}{2}$$
If $p$ is odd, then $p+1$ is even, and this integer factors into $p$ and $\frac{p+1}{2}$. Any prime different than $p$ cannot divide $p$, so if each prime $q$ that's less than $p$ divides the number $p\cdot (\frac{p+1}{2})$, then they must all divide the number $\frac{p+1}{2}$.
But there is always a prime $q$ such that $(\frac{p+1}{2})<q<2(\frac{p+1}{2})=p+1$ when $\frac{p+1}{2}>1$, by Bertrand's postulate (Wikipedia link), and such a prime cannot go into $\frac{p+1}{2}$ since it is bigger than it. Such a prime is also less than $p$ unless $p=q=2$. Moreover, $\frac{p+1}{2}>1$ for any odd prime $p$.
Thus, the only possibilities are $p=2$ and $p=3$, and it can be easily  checked that the statement is valid only for $p=3$. (For $p=2$ we have $2\nmid 3$, and for $p=3$, we have $2\mid 6$ and $3\mid 6$.) 
A: Let $p_n$ be the $n$-th prime.
By Betrand's Postulate, $p_n < 2^n$ for $n > 1$ and therefore $S_n:=\dfrac{p_n(p_n+1)}{2} < 2^{2n-1} + 2^{n-1} < 2^{2n} = 4^n$.
But $5 < p_n$ for $n > 3$ so $P_n:=\displaystyle\prod_{i=1}^n p_i > 5^{n-1}$ for $n > 3$.
Since $P_n$ must divide $S_n$ then $P_n \leq S_n$, so either $5^{n-1} < 4^n$ and $n > 3$ or $n\leq 3$.
The first case means that $n < 8$ since $5^7 > 4^8$ and thus $5^{n-1} > 4^n$ for $n\geq 8$.
$p_1 = 2$ and $S_1 = 3$ but $2 \leq 2$ and $2\nmid S_1$.
$p_2 = 3$ and $S_2 = 6$ is divisible by $P_2 = 6$. So $p=3$ is a solution.
$p_3 = 5$ and $S_3 = 15$ but $2 < 5$ and $2\nmid S_3$.
$p_4 = 7$ and $S_4 = 28$ but $3 < 7$ and $3\nmid S_4$.
$p_5 = 11$ and $S_5 = 66$ but $7 < 11$ and $7\nmid S_5$.
$p_6 = 13$ and $S_6 = 91$ but $3 < 13$ and $3\nmid S_6$.
$p_7 = 17$ and $S_7 = 153$ but $2 < 17$ and $2\nmid S_7$.
The only solution for $p$ is $3$.
A: Let $p_n$ denote the $n$-th prime. Note that $1+2+\cdots+p_n = p_n\cdot\dfrac{p_n+1}{2}$. 
By Bertrand's postulate, $p_n < 2p_{n-1}$. Hence, $0 < \dfrac{p_n+1}{2} < \dfrac{2p_{n-1}+1}{2} < 2p_{n-1}$. 
Therefore, $\dfrac{p_n+1}{2}$ is not divisible by $2p_{n-1}$. 
If $n \ge 3$, then $2$ and $p_{n-1}$ are distinct primes, and so $\dfrac{p_n+1}{2}$ is not divisible by both $2$ and $p_{n-1}$. 
Then, since $p_n$ is distinct from $2$ and $p_{n-1}$, we have $p_n\cdot\dfrac{p_n+1}{2}$ is not divisible by both $2$ and $p_{n-1}$. 
Hence, $1+2+\cdots+p_n$ is not divisible by all of $p_1 = 2, \ldots, p_{n-1}, p_n$ for $n \ge 3$. 
For $n = 1$, we have $p_1 = 2$, and $1+2 = 3$ is not divisible by $p_1 = 2$.
For $n = 2$, we have $p_2 = 3$, and $1+2+3 = 6$ is divisible by $p_1 = 2$ and $p_2 = 3$.
Therefore, the only prime $p$ such that $1+2+\cdots+p$ is divisible by all primes less than or equal to $p$ is $p = 3$.
A: The obvious way to start would be to check the results for a few small numbers and review the patterns discovered.
For any odd prime $p$ the sum of numbers $s$ from $1$ to $p$ is divisible by $p$ : $s=\frac 12 p(p+1)$ and $2\nmid p$ (or for a more intuitive reason, you can form the numbers below $p$ in separate pairs that sum to $p$). By examination we can see that $2 \nmid (1+2)$.
Looking at this sum, it is clear that all the primes below $p$ need to divide $\frac{p+1}{2}$. However for primes $7$ and greater, the product of all the smaller primes is greater than the prime concerned, so we only need to consider $p=3$ and $p=5$. For $p=5$, $\frac{p+1}{2}$ is odd so this is eliminated and only $p=3$ fits the criteria.
