[spoiler]Okay, we need to find the total number of 0's at the end of 30!. This is how you do so:

We can make a 10 out of each factor of 10 less than or equal to 30. There are 3 of these.

We can make extra factors of 10 by putting 2's and 5's together. We have: (2, 5) (12, 15) (22, 25). HOWEVER, note that 25 has TWO fives in it... we can put this together with an 8 and make yet ANOTHER 10. This is all we can do.

So, we have that there are 7 factors of 10 in 30!, and 30! ends in 7 zeros. From this, we see that (1) isn't sufficient, but if d > 6, then d has to be 7. So both together are sufficient. So, C.[/spoiler]

Tricky but I will give you a clue: 25 is 5*5

Wow that's a brilliant solution.
What percentile do you think this question is (i.e., what level)?

Ok here is an easier way. 10 is because of 2 and 5. The number of 5s are o course greater than number of 2s. therefore number of 0s will be determined by number of 5s. Simply calculate the number of 5 by the following:-

[x] : denotes the greatest integer less than x. i.e [3.2] = 3

Number of 5s = [30! / 5] + [30! / 5^2] + [30! / 5^3] .... = 6 + 1 = 7. This is only an easier way of calculating number of 0s in a given number.