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.
Watch the lessons below for more detailed explanations of the concepts tested in this question.
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