GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 16 Oct 2019, 23:41

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

In the expression 13!/x^3, for which of the following values of x will

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58403
In the expression 13!/x^3, for which of the following values of x will  [#permalink]

Show Tags

New post 23 Jun 2017, 03:34
2
1
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

86% (01:25) correct 14% (01:50) wrong based on 85 sessions

HideShow timer Statistics

Senior PS Moderator
User avatar
V
Joined: 26 Feb 2016
Posts: 3340
Location: India
GPA: 3.12
In the expression 13!/x^3, for which of the following values of x will  [#permalink]

Show Tags

New post 23 Jun 2017, 03:51
\(13! = 13*12*11*10*9*8*7*6*5*4*3*2*1 = 13*12*11*7*5^2*3^4*2^8\)

Of the answer options for values of x, at x=5
the expression \(\frac{13!}{x^3}\) will not be an integer.

Hence Option C is the correct answer.
_________________
You've got what it takes, but it will take everything you've got
Senior SC Moderator
avatar
V
Joined: 22 May 2016
Posts: 3548
In the expression 13!/x^3, for which of the following values of x will  [#permalink]

Show Tags

New post 23 Jun 2017, 08:28
Bunuel wrote:
In the expression \(\frac{13!}{x^3}\), for which of the following values of x will the expression will NOT be an integer?

A. 3
B. 4
C. 5
D. 6
E. 8

Method I - Raise each answer choice to the power of 3 because we're looking for \(x^3.\). Prime factorize each answer choice if not prime.

Then find the number of powers of the prime(s) in 13! The answer will be an integer if and only if there are enough powers of the prime factor(s) in the factorial. Using formula:

\(\frac{13}{n} + \frac{13}{n^2} + \frac{13}{n^3}\) . . . only to the point where n raised to whatever power is ≤ 13.**

A. \(3^3\). We need three powers of 3.

\(\frac{13}{3^1}= 4\)

\(\frac{13}{3^2} = 1\)

13 cannot be divided by \(3^3\) to get an integer, so stop there.

There are 4 + 1 = 5 powers of 3 in 13!. We only need 3. \(\frac{13!}{3^3}\) will be an integer.

B. \(4^3 = (2^2)^3 = 2^6\). We need six powers of 2.

\(\frac{13}{2}\) = 6

\(\frac{13}{2^2}\) = 3

\(\frac{13}{2^3 =
1}\)
, so

There are 6 + 3 + 1 = 10 powers of 2 in 13! We need six powers of 2 to make \(4^3\). We have ten. \(\frac{13!}{4^3}\) will be an integer.

C. \(5^3\)

\(\frac{13}{5} = 2\)

13 cannot be divided by \(5^2\) to get an integer.

We only have two powers of 5 and we need three. \(\frac{13!}{5^3}\) will not be an integer.

D. \(6^3 = 2^3*3^3\). Check to see that there are enough powers of both 2 and 3.

From answer B, there are ten powers of 2, and we need only three.

From answer A, there are five powers of 3, also enough. \(\frac{13!}{6^3}\) will be an integer.

E.\(8^3 = (2^3)^3\), or \(2^9\)

From Answer B: there are ten powers of 2, and we need only nine. \(\frac{13!}{8^3}\)will be an integer.

Method II - Look at the factors of 13! and see whether or not for each answer choice there are three powers of the number in 13!

13! = 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1.

A.\(3^3\). Number of powers of 3 in 13!
3 has one three
6 has one three
9 has two threes
12 has one three
Total of five powers of 3. We only need three to get even division and thereby an integer. \(\frac{13!}{3^3}\) will be an integer.

B. \(4^3\). Either prime factorize 4, or notice:
4 has one four
8 has two fours
12 has one four. We have the necessary three powers of 4. \(\frac{13!}{4^3}\) will be an integer.

C. \(5^3\)
5 has one five
10 has one five
There are no other powers of 5. We have two, we need three. \(fraction]13!/5^3\)[/fraction] will not be an integer.

D. \(6^3\) = \(2^3*3^3\). There aren't enough obvious powers of 6 in 13! There is one in 6, and one in 12. We need three powers of 6.

But by prime factorizing \(6^3\) to \(2^33^3\)

From answer A, there are five powers of 3.
We can get one power of 2 from 2, and two powers of 2 from 4, total three as needed.

There are enough powers of 2 and 3 that \(\frac{13!}{6^3}\) will be an integer.

E.\(8^3 = (2^3)^3 = 2^9\)
2 has one two
4 has two twos
6 has one two
8 has three twos
10 has one two
12 has two twos. That's 10 total. We need 9.\(\frac{13!}{8^3}\) will be an integer.

Answer:
(C)

** See Bunuel 's excellent Math Number Theory, here https://gmatclub.com/forum/math-number-theory-88376.html, and scroll down to Factorials, Finding the number of powers of a prime number \(p\), in the \(n!\)
_________________
SC Butler has resumed! Get two SC questions to practice, whose links you can find by date, here.




Choose life.
SVP
SVP
avatar
B
Joined: 06 Nov 2014
Posts: 1873
Re: In the expression 13!/x^3, for which of the following values of x will  [#permalink]

Show Tags

New post 23 Jun 2017, 10:37
Bunuel wrote:
In the expression \(\frac{13!}{x^3}\), for which of the following values of x will the expression will NOT be an integer?

A. 3
B. 4
C. 5
D. 6
E. 8


Since we are dividing by 3rd power of x.
Hence there should be atleast 3 powers of x in the numerator:

x = 3 -> 13! has more than 3 powers of 3. Integer
x = 4 -> 13! has more than 3 powers of 4. Integer
x = 5 -> 13! has 2 powers of 5. Not an integer.

Correct Option C
Retired Moderator
avatar
P
Joined: 22 Aug 2013
Posts: 1429
Location: India
Re: In the expression 13!/x^3, for which of the following values of x will  [#permalink]

Show Tags

New post 23 Jun 2017, 10:42
13! = 13*12*11*10...*5*4..*2*1

The number of 5's in 13! are only '2': one in '10' and the other in '5'. So 13! will NOT be divisible by 5^3.

Hence C answer
Target Test Prep Representative
User avatar
G
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2816
Re: In the expression 13!/x^3, for which of the following values of x will  [#permalink]

Show Tags

New post 26 Jun 2017, 17:57
Bunuel wrote:
In the expression \(\frac{13!}{x^3}\), for which of the following values of x will the expression will NOT be an integer?

A. 3
B. 4
C. 5
D. 6
E. 8


Let’s prime factorize 13!.

13! = 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2

13! = 13 x (2^2 x 3^1) x 11 x (5 x 2) x 3^2 x 2^3 x 7 x (2 x 3) x 5 x 2^2 x 3 x 2

13! = 13 x 11 x 7 x 5^2 x 3^5 x 2^10

Let’s now analyze each answer choice:

A) x = 3

Thus, x^3 = 3^3.

Since there are five 3s in 13!, x can be 3.

B) x = 4 = 2^2

Thus, x^3 = 2^6.

Since there are ten 2s in 13!, x can be 4.

C) x = 5

Thus, x^3 = 5^3.

Since there are only two 5s in 13!, x cannot be 5.

Answer: C
_________________

Jeffrey Miller

Head of GMAT Instruction

Jeff@TargetTestPrep.com
TTP - Target Test Prep Logo
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

VP
VP
User avatar
D
Status: Learning
Joined: 20 Dec 2015
Posts: 1007
Location: India
Concentration: Operations, Marketing
GMAT 1: 670 Q48 V36
GRE 1: Q157 V157
GPA: 3.4
WE: Engineering (Manufacturing)
Reviews Badge
Re: In the expression 13!/x^3, for which of the following values of x will  [#permalink]

Show Tags

New post 27 Jun 2017, 10:49
Imo C
The power of 3 in 13!=5
The power of 4 in 13!=3
The power of 5 in 13!=2
The power of 6 in 13!=2
The power of 8 in 13!=1
The power of 2 in 13!=10
So 3! will be divisible by 27 ,64,512,216 but will not be divisible by 125 so C is the answer.
_________________
Please give kudos if you find my answers useful
Retired Moderator
User avatar
P
Joined: 19 Mar 2014
Posts: 922
Location: India
Concentration: Finance, Entrepreneurship
GPA: 3.5
GMAT ToolKit User
Re: In the expression 13!/x^3, for which of the following values of x will  [#permalink]

Show Tags

New post 27 Jun 2017, 12:36
1
In the expression \(\frac{13!}{x^3}\), for which of the following values of x will the expression will NOT be an integer?

In order to find what values of x will NOT be an integer, lets first find out how are we going to get the integer.

So, in order to get an integer we want denominator to completely divide the numerator, in other words can we change the denominator to 1

This is only possible if there are sufficient factors which can convert all the three units of denominator numbers to 1

Lets evaluate the answer choices, if they have at least three factors in order to completely divide the numerator.

A. 3 =====> Factors in numerator - 3, 6 & 9 ===> Gives Integer
B. 4 =====> Factors in numerator - 4, 8 & 12 ===> Gives Integer
C. 5 =====> Factors in numerator = 5 & 10 ===> Does not Give Integer as Third Unit of denominator cannot completely divide the numerator.
D. 6 =====> Factors in numerator = 2*3, 6, 12 ===> Gives Integer
E. 8 =====> Factors in numerator = 2*4, 8,4 (From 12), 2 (From 10), 2 (From 6) {4*2*2 = 8} ===> Gives Integer

Hence, Answer is C

You Liked the Solution? Appreciate by giving Kudos :good
_________________
"Nothing in this world can take the place of persistence. Talent will not: nothing is more common than unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not: the world is full of educated derelicts. Persistence and determination alone are omnipotent."

Best AWA Template: https://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html#p470475
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 13204
Re: In the expression 13!/x^3, for which of the following values of x will  [#permalink]

Show Tags

New post 19 Jan 2019, 04:17
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Club Bot
Re: In the expression 13!/x^3, for which of the following values of x will   [#permalink] 19 Jan 2019, 04:17
Display posts from previous: Sort by

In the expression 13!/x^3, for which of the following values of x will

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne