Partial sums of primes $2+3+5+7+11+13...$ is clearly the sum of the primes.
Now I consider partial sums such:
$2+3+5+7+11=28$ which is divisible by $7$
My question is:
are there infinitely many partial sums such that:
$p_1+p_2+p_3+...+p_{k}+p_{k+1}=m*p_{k}?$ with $m$ some positive integer? With Pari/gp apparently up to 10^10 there are only two examples $7$=$p_k$ and $8263=p_k$. Heuristically do you think that infinitely many such partial sums should exist? Note: 7 and 8263 are both primes belonging to primes on the left side of the triangle formed by listing successively the prime numbers in a triangular grid. See https://oeis.org/A078721
Note in both cases $2+3+5+7=17$ is prime and $2+3+5+...+p_{1036}=3974497$ is prime. I note that $17$ and $3974497$ are primes of the form $4s+1$, whereas $p_4=7$ and $p_{1036}=8263$ are primes of the form $6s+1$.
$7$ and $8263$ are primes such that starting from the right, the odd positioned digits are prime and the even positioned digits are composite. But also $5$ and $8243$ which are the previous primes have this property. No other prime of this type found below $10^{12}$
I noticed that 7! has 4 digits where 4 is a palindrome. 8263! has 28782 digits where 28782 is a palindrome.
 A: You asked for a heuristic answer.
There is an heuristic argument that infinitely many such partial sums should exist.  Consider $P(k)$, an heuristic estimate of the probability that the partial sum of the first $k+1$ primes would be divisible by $p_k$.  Now $$p_k \sim k \log k$$ and if only random chance were involved,  $$P(k) \approx \frac1{p_k}  \sim \frac1{k \log k}$$ 
In that case, the expected number of primes with the property you want would be something like
$$\int_2^\infty \frac1{x \log x}\,dx$$
and that integral diverges to infinity.
The reason it seems so rare is that the rate of divergence is like $\log(\log x)$ and while that function goes to infinity, "nobody ever sees it do so."
On the other hand, proving that there an infinite number of such values of $k$ (in the same sense that Euclid's argument proves there is no last prime) is probably quite difficult.  And if the conjecture that there are an infinite number of such values of $k$ turned out to be false,  proving that some particular $k$ is the last one with this property would seem to be even harder.  
