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Greatest integer k?

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by crackgmat007 » Sun May 10, 2009 7:24 pm
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?
A. 10
B. 12
C. 14
D. 16
E. 18

IMO - A, product of numbers from 1 to 30 will be having 10 multiples of 3, hence answer must be A. Am I missing something? The answer seems to be C
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by sureshbala » Sun May 10, 2009 8:01 pm
We need to find the highest power of 3 in 30! and the procedure is as follows...

Divide 30 successively by 3 and consider only the quotients i.e

30/3 = 10

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

3/3 = 1

Now add all these quotients...

Hence the highest power of 3 in 30! is 10+3+1 = 14.

Thus the max value of k = 14
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by crackgmat007 » Sun May 10, 2009 8:47 pm
Seems pretty simple. Thanks for solving. Curious to know the logic behind this.
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by sureshbala » Sun May 10, 2009 8:55 pm
The logic is simple.

In 30!, we see 10 multiples of 3

i.e. 3, 6, 9, 12, 15, 18, 21, 24, 27 and 30.

30/3 = 10 represents the same.

In the above multiples of 3, there are multiples which contain more than one 3 i.e. we have multiples of 9

9, 18, 27

10/3 = 3 represents these three multiples.

These three multiples contain two 3's in them ...one 3 is counted in the first round and the other 3 is counted in the second round.

Again, there could be three 3's i.e. we need multiple of 27.

Only 27 in the given range

3/3 = 1 represents this.


So the three 3's in 27 are counted one each in the above three rounds.

Thus this continues

Next we may have four 3's i.e. a multiple of 81.

But 81 is out the given range.

Hence highest power of 3 in 30! = 10 + 3 + 1 = 14
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