Video Explanation

Text Explanation

Let's begin by examining 15! It is the product of all integers from 1 to 15: 

This question is essentially asking us the following: "what is the greatest number of 3s that we can factor out of this product. 

Now notice that in addition to the 3 in the product above, there are several numbers that have 3s "hidden" in them. In fact, all of the numbers in the product that are multiples of 3 have 3s "hiding" in them. These numbers are the following: 

6 = 2  3  

9 = 3  3

12 = 3 4 

15 = 3  5 

So the numbers 6, 12, and 15 have one 3 "hiding" in them, and the number 9 has two 3s "hiding" in it. 

We can rewrite 15! so that all of these 3s are shown: 

Now notice that we have a total of six 3s in the above product. We can factor out these 3s as follows: 

Now we know that  must be a divisor of 15! Six is the greatest number of 3s that we can factor out of 15! 

Keep in mind that the question told us that i s a divisor of 15!, and that k is the greatest integer possible such that  will be a divisor of 15!. This means that k must be equal 6. So the right answer is D.