OG 12 # 110 Factors

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OG 12 # 110 Factors

by EMAN » Wed Sep 09, 2009 6:42 pm
Hello. I'm trying to understand the explanation for this problem here.

Problem 110 in the OG 12 states:

If p is the product of the integers from 1 to 30 inclusive, what is the greatest integer k for which 3^k is a factor of p?

Answer:

3
6 = 2 x 3
9 = 3 x 3
12 = 2 x 2 x 3
15 = 3 x 5
18 = 2 x 3 x 3
21 = 3 x 7
24 = 2 x 2 x 2 x 3
27 = 3 x 3 x 3
30 = 2 x 3 x 5

Therefore the answer is 3^14 because there are 14 factors of 3 above. What principle states that you can multiply 3 to the power of the number of factors? Is there any other way to solve this problem that's easier? Any assistance is greatly appreciated.
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by sureshbala » Wed Sep 09, 2009 9:57 pm
Hi,

Here is the quickest way to find highest power of a prime number in any factorial.

In this question we need to find the highest power of 3 in 30!.

Divide 30 successively by 3 and add the respective quotients.

30/3 = 10

10/3 = 3 (don't worry about remainder)

3/3 = 1

Now add 10+ 3 + 1 = 14

Thus the highest power of 3 in 30! = 14

(Will try to explain the logic when the time permits)

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Got it

by EMAN » Thu Sep 10, 2009 5:38 am
Wow, that's significantly better than the OG explanation. I would be curious as to the logic. Thanks!