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# If p is the product of the integers from 1 to 30, inclusive,

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If p is the product of the integers from 1 to 30, inclusive, [#permalink]

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17 Jul 2008, 01:55
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Show the quickest way to solve this.
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Jimmy Low, Frankfurt, Germany
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Re: Math Set 1: Q15 [#permalink]

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17 Jul 2008, 02:58
jimmylow wrote:
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

Show the quickest way to solve this.

C

Since 3 is a prime number, the numbers 1 to 30 that are relevant here are only numbers with 3 as a factor, i.e., multiples of 3:

3, 6, 9...27, 30

3 numbers have 3 as a factor multiple times: 9, 18, 27. 7 have 3 as a factor only once.
9 = 3^2
18 = 2*3^2
27 = 3^3

Counting up:
7+2+2+3 = 14
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Re: Math Set 1: Q15 [#permalink]

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06 Aug 2008, 16:52
jimmylow wrote:
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

Show the quickest way to solve this.

Fastest way to solve:

30/3 = 10
10/3 = 3 (with a remainder)
3/3 = 1

Then just add up the quotients: 10 + 3 + 1 = 14
C
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Re: Math Set 1: Q15 [#permalink]

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06 Aug 2008, 18:12
jimmylow wrote:
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

Show the quickest way to solve this.

3,6,9,12,15,18,21,24,27,30 -------> total is 14 => k=14 can be the highest power of 3 to make 3^k a divisor of p
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Re: Math Set 1: Q15   [#permalink] 06 Aug 2008, 18:12
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