If k and n are both integer and k > n > 0 is k!/n! divisible by 30 ?

Analyzing the question:
30 = 2 * 3 * 5. We need a factor of 2, 3, and 5 from (k!/n!). If k and n have a difference of at least 5, we automatically get these factors from the numbers between k and n. For example, if k = 13 and n = 8, then we have k! / n! = 13 * 12 * 11 * 10 * 9 which is 5 numbers multiplied since k - n = 5. We can find a multiple of 5 for sure since there are 5 consecutive numbers multiplied, a multiple of 2 and 3 as well.

Statement 1:
If k and n are too far apart (k!/n!) will have a factor of 30 as demonstrated above. So let us focus on the closer k and n pairs, k = 16 and n = 14 -> (k!/n!) = 16 * 15. It has a factor of 30. If we take a bigger k we can see it will expand the factors for (k!/n!) so this is sufficient. (e.g. k = 17 n = 13, k!/n! = 17*16*15*14 which is more factors)

Statement2:
As demonstrated above, this is sufficient.

Ans: D