For any prime number p, there exists a positive integer n such that the sum of the first n consecutive odd prime numbers is divisible by p (let p be a prime number).
p | (3 + 5 + 7 + ... + P_n)
where P_n is the nth odd prime number and "|" denotes divisibility.
Possible approach for a proof:
Define a sequence of odd prime numbers P_1, P_2, P_3, ..., P_n, where P_1 = 3, P_2 = 5, P_3 = 7, and so on. Compute partial sums of the sequence: S_i = P_1 + P_2 + ... + P_i for i = 1, 2, 3, ..., n.
Intent: there exists an i such that p | S_i. If p = 3, then S_1 = 3, so p | S_1. In this case, the conjecture holds. If p > 3, consider the sequence of residues (remainders) when each S_i is divided by p: R_i = S_i mod p, for i = 1, 2, 3, ..., n. There are p-1 possible residues, from 1 to p-1, and at most n of them in our sequence. If two partial sums, S_i and S_j, have the same residue (i.e., R_i = R_j), then p divides the difference between the sums (S_j - S_i = P_{i+1} + ... + P_j). In this case, the conjecture holds for some i. If no two partial sums have the same residue, then there must be a partial sum S_i such that R_i = 0 (i.e., p | S_i). In this case, the conjecture holds for i.
In all cases, the conjecture holds for some positive integer n.