A box contains bags of marbles. All of the bags hold the

Following would be your chain of thought:

All bags have the same number of marbles except 1 bag which has 1 marble extra. There are 2001 marbles. If you discard that one extra marble, 2000 marbles must be equally distributed among n bags. So number of bags must be a factor of 2000 e.g. if there are 200 bags, each bag will have 10 marbles now. If there are 20 bags, each bag will have 100 marbles now etc. Basically, 2000 is to be split up into two factors: number of bags and number of marbles.

\(2000 = 2^4 * 5^3\)

(1) The number of bags is between 13 and 23 inclusive
No of bags could be 16 (in which case each bag has 125 marbles except one bag with 126 marbles)
No of bags could be 20 (in which case each bag has 100 marbles except one bag with 101 marbles)
There are no other factors of 2000 between 13 and 23.
Not sufficient.

(2) There is an even number of bags, and there is an even number of marbles in the bag containing the extra marble
The bag with the extra marble has even number of marbles so that bags with the same number of marbles must have an odd number of marbles.
If number of marbles are odd, when we split 2000 in 2 factors, all 2s should go in the 'number of bags' factor.
No of bags could be 16 (in which case each bag has 125 marbles except one bag with 126 marbles)
No of bags could be 16*5 (in which case each bag has 25 marbles except one bag with 26 marbles)
No of bags could be 16*5*5 (in which case each bag has 5 marbles except one bag with 6 marbles)
Not sufficient.

Using both, we see that only 16 is common between the statements. Hence, number of bags must be 16.

Answer (C)