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by lisannemuns » Tue Oct 01, 2013 12:19 am
Hello, Can somebody help me with this question?

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

a)10
b)12
c) 14
d) 16
e) 18
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by Brent@GMATPrepNow » Tue Oct 01, 2013 12:27 am
lisannemuns wrote:Hello, Can somebody help me with this question?

If [p] is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for wich 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: C

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by GMATGuruNY » Tue Oct 01, 2013 2:24 am
lisannemuns wrote:Hello, Can somebody help me with this question?

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

a)10
b)12
c) 14
d) 16
e) 18
[p] = 30!.
3^k = the number of 3's that can be divided into [p].
A very fast approach is to count how many times EACH POWER OF 3 can be divided into 30:

30/3¹ = 10.
The calculation above indicates that 30! includes 10 multiples of 3¹.

30/3² = 30/9 = 3.
The calculation above indicates that 30! includes 3 multiples of 3².

30/3³ = 30/27 = 1.
The calculation above indicates that 30! includes 1 multiple of 3³.

Thus, [p] includes 10 multiples of 3, 3 multiples of 3², and 1 multiple of 3³.
Thus, the total number of 3's that can be divided into [p] = 10+3+1 = 14.

The correct answer is C.
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by [email protected] » Tue Oct 01, 2013 12:14 pm
Hi lisannemuns,

Both Brent and Mitch have properly explained a straight-forward method and the "math" behind the correct answer. For future reference, in these types of situations, it's quite common for there to be "more of what you're looking for than you originally think exist."

In this question, we're looking for all of the 3s.

Most people would just could the "multiples" of 3. Here, we'd have 10 multiples of 3; many people would choose answer A and not realize that they were wrong. With a long enough string of values, you'll likely come across values that have MORE THAN ONE 3. Here, the values 9 and 18 each have two 3s, while the value 27 has three 3s. Be on the look out for this "twist" since it's likely to occur in this type of question.

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by theCodeToGMAT » Wed Oct 02, 2013 9:23 am
Divide 30 by 3

SO,

30/3 = 10
10/3 = 3
3/3= 1

10+3+1 = 14

Answer [spoiler]{C}[/spoiler]
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