If an alien were abducted by humans, and it memorized a much larger prime number, what would we give in exchange for that piece of information?
So, nothing. The alien could say "Oh yeah, we have 2^34240923842043983204982-1" and they could be absolutely correct but there's nothing we could do to know.
Or, if their number was big but not that big, it might be able to be verified in, say, a year, in which case it might win the EFF's prize and could possibly get some other sort of prize money. So, $150k + some undefined amount.
The proof is left as an exercise for the student.
To check, 2^6-1=63 is divisible by three (it might work), but 2^9-1 has a sum of 12 and a value of 511, which is not divisible by three (it's 170 * 3 = 510 + 1).
That's how you solve all problems of this form. For any a and b, a^n mod b eventually becomes periodic as n grows. It's easy to prove because there's only a finite number of possible remainders mod b. As an exercise, calculate 3^1000000 mod 7.
3 is prime. 34240923842043983204982 is divisible by 3 - 1 = 2.
ab mod n = [a mod n * b mod n] mod n
So, we can automatically infer that 2 ^ 34240923842043983204982 ~= 1 mod 3.
After subtracting 1, it will be divisible by 3.
Alternatively, you could notice that 2 ^ 2 ~= 1 mod 3, which implies 2 ^ 4 ~= 1, 2 ^ 6 ~= 1, so 2 ^ all even powers will be congruent to 1 modulo 3.
So 2 ^ x - 1 will always be divisible by 3 for even x.
Everybody contributes to the MP for free but that doesn't mean there isn't any value involved. It's a donation of sorts.