A box of bags containing marbles

Nice work, the answer is C

Here's my solution as well:

Assume that we have b bags and almost all bags contain m marbles (except the one bag that has m+1 marbles)

We know that bm + 1 = 2001
Or we can say that bm = 2000
It will help with this question to find the prime factorization of 2000.
We get 2000 = 2x2x2x2x5x5x5

Now we can examine the statements
(1) We need to write bm = 2000 such that b is between 13 and 23. We can see that b can equal 16 (as in b=16 and m=125), or b can equal 20 (as in b=20 and m=100) INSUFF

(2) Another way to say that there is an even number of marbles in the bag containing the extra marble is to say that there is an odd number of marbles in all of the other bags. In other words, m is odd.
How many ways are there to split 2x2x2x2x5x5x5 into two parts such that one part is even and the other part is odd? Several. We can have b=16 and m=125, and we can have b=80 and m=25). INSUFF

(1) and (2): Statement (1) tells us that either b=16 (and m=125) or b=20 (and m=100). Statement (2) tells us that m must be odd. So, the only conclusion is that b=16 and m=125. SUFF