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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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20 Jun 2010, 23:59
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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
[Reveal] Spoiler: OA
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Re: OG12 - PS [#permalink]

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21 Jun 2010, 02:23
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divakarbio7 wrote:
If p is the product of the integers from 1 to 30,
inclusive, what is the greatest integer k for which 3k is
a factor of p ?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18

Not sure as to how to approach this problem

Given: $$p=30!=3^k*n$$. Question: $$k_{max}=?$$.

We should find highest power of 3 in 30!. Finding the power of a prime in n!: everything-about-factorials-on-the-gmat-85592.html or math-number-theory-88376.html (factorials chapter).

So:

$$\frac{30}{3}+\frac{30}{9}+\frac{30}{27}=10+3+1=14$$.

Answer: C.
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Re: official guide 13th ed #116 problem solving [#permalink]

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19 Jun 2012, 06:13
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peter2000 wrote:
p is product of integers from 1 to 30 inclusive, what is the greatest integer k for which3^k is a factor of p?

A)10
B)12
C)14
D)16
E)18

I understand the way the GMAT guide does the problem, I was wondering if there was a quicker way?

Check out this post for an explanation of the method you can use to solve such questions:
http://www.veritasprep.com/blog/2011/06 ... actorials/
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23 Jun 2010, 13:38
15. No need to actually find the product of 1 to 30. Just look at the numbers that have factors of 3.

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

There are 14 3's, so the largest factor of p for 3k is 3*14.
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26 Feb 2012, 11:19
$$\frac{30}{3}+\frac{30}{3^2}+\frac{30}{3^3}$$
= 10 + 3 + 1 (ignore the decimals part of each fraction)
= 14
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Re: If p is the product of the integers from 1 to 30, inclusive [#permalink]

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01 Oct 2012, 06:50
P = 30!
8
P = 30 x 29 x 28 x 27 x 26 x 25 x 24 x 24 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 09 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

Out of these 30 , 27 , 24 , 21 , 18, 15 , 12 , 09 , 06 , 3 are factors of 3

3 x 10 , 3 x 3 x 3 , 3 x 8 , 3 x 3 x 2, 3 x 5 , 3 x 4 , 3 x 3 x 3 , 3 x 2 , 3

So we have a total of 14 three's ...

Therefore the maximum value of K can be 14 (C)
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Re: If p is the product of the integers from 1 to 30, inclusive [#permalink]

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

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

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16 Aug 2016, 15:56
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divakarbio7 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

This question is really asking us to determine the number of 3's "hiding" in the prime factorization of p.

p = (1)(2)(3)(4)(5)(6)(7)(8)(9) . . . (27)(28)(29)(30)
= (1)(2)(3)(4)(5)(2)(3)(7)(8)(3)(3)(10)(11)(3)(4)(13)(14)(3)(5)(16)(17)(3)(3)(2)(19)(20)(3)(7)(22)(23)(3)(8)(25)(26). . . (3)(3)(3)(28)(29)(3)(10)
= (3)^14(other non-3 stuff)

Answer:
[Reveal] Spoiler:
C

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

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24 Jan 2017, 19:52
1) We need to calculate the number of prime factors 3 in the product of all the integers from 1 to 30
2) 30/3=10; 30/9=3; 30/27=1; 10+3+1=14

The correct answer is C.
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Re: If p is the product of the integers from 1 to 30, inclusive [#permalink]

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25 Mar 2017, 10:20
30/3, 30/9, 30/27
10+3+1=14
Answer is C
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Re: If p is the product of the integers from 1 to 30, inclusive [#permalink]

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26 Mar 2017, 00:35
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

product of 1 to 30 includes the multiples of 3 and the number of 3 appear ,count them and add

3-1
6-1
9-2
12-1
15-1
18-2
21-1
24-1
27-3
30-1
=14
Re: If p is the product of the integers from 1 to 30, inclusive   [#permalink] 26 Mar 2017, 00:35
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# If p is the product of the integers from 1 to 30, inclusive

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