If you were famous....

[awpertunity]

They are used to classify how "difficult" a problem is, where checking all possible answers for a solution is not feasible.

A P problem is a problem that can be solved using a computer/algorithm in polynomial time.
An NP problem is a problem whose solution can be checked in polynomial time.

So here's a problem example: if I give you a finite set of numbers {-10, -5, 2, 4, 9, 19}, can you find a subset of them that sum to 0?

This problem is clearly NP because if I give you a solution (or possible solution), it is very easy to check if that solution satisfies the condition. i.e., {-10, -5, 2, 4, 9} is a solution and can be checked by simply seeing -10 - 5 + 2 + 4 + 9 = 0. Similarly you can check any possible solution, for instance, {-10, 19} is not a solution by seeing -10 + 19 =/= 0.

So the question is, does P = NP, meaning if a problem is NP, is it P? So if there is a way to check a solution in polynomial time, is there a way to actually find a solution in polynomial time. Checking every possible combination is clearly a way to find a solution, but cannot be done in polynomial time. As the size of the set I give you grows, the number of combinations you need to check grows exponentially rather than polynomially. But if P=NP there would mean there is some smarter algorithm that would be able to find a solution in polynomial time.

Here's the $1,000,000 question at hand http://www.claymath.org/millennium/P_vs_NP/ as well as the 6 other millennium problems if you're feeling ambitious

I believe the Poincare conjecture has already been solved...