<s>[INST] 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. 
 [/INST] 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 $p = 3$.
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