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

It is currently 20 Jun 2019, 12:35

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

D01-16

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

Hide Tags

 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55732
D01-16  [#permalink]

Show Tags

New post 16 Sep 2014, 00:12
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

68% (00:55) correct 32% (00:32) wrong based on 161 sessions

HideShow timer Statistics

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55732
Re D01-16  [#permalink]

Show Tags

New post 16 Sep 2014, 00:12
1
Official Solution:


If \((|p|!)^p = |p|!\), which of the following could be true?

I. \(p=-1\)

II. \(p=0\)

III. \(p=1\)


A. I only
B. II only
C. III only
D. II and III only
E. I, II and III


Important properties: \(0!=1\) and any non-zero number to the power of 0 is 1.

Let's check the options:

If \(p=-1\) then \((|p|!)^p = (|-1|!)^{-1}=1^{-1}=1\) and \(|p|!=|-1|!=1!=1\) so \(p\) could be -1;

If \(p=0\) then \((|p|!)^p = (|0|!)^{0}=1^{0}=1\) and \(|0|!=0!=1\) so \(p\) could be 0;

If \(p=1\) then \((|p|!)^p = (|1|!)^{1}=1^{1}=1\) and \(|p|!=|1|!=1\) so \(p\) could be 1.


Answer: E
_________________
Intern
Intern
User avatar
Joined: 28 Oct 2011
Posts: 6
Location: Qatar
Concentration: International Business, Marketing
Schools: HBS '17
GMAT 1: 510 Q50 V26
GPA: 2.68
WE: Marketing (Energy and Utilities)
GMAT ToolKit User
Re: D01-16  [#permalink]

Show Tags

New post 11 Dec 2014, 06:11
1^-1 = 1 ? Please explain.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55732
Re: D01-16  [#permalink]

Show Tags

New post 11 Dec 2014, 06:30
Ram1987 wrote:
1^-1 = 1 ? Please explain.


Negative powers:
\(a^{-n}=\frac{1}{a^n}\), hence \(1^{(-1)}=\frac{1}{1^1}=1\).

Theory on Exponents: math-number-theory-88376.html
Tips on Exponents: exponents-and-roots-on-the-gmat-tips-and-hints-174993.html

All DS Exponents questions to practice: search.php?search_id=tag&tag_id=39
All PS Exponents questions to practice: search.php?search_id=tag&tag_id=60

Tough and tricky DS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125967.html
Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html

_________________
Intern
Intern
avatar
Joined: 27 Oct 2014
Posts: 23
GMAT 1: 680 Q49 V34
GPA: 3.8
GMAT ToolKit User Reviews Badge
Re: D01-16  [#permalink]

Show Tags

New post 11 Dec 2014, 09:00
It seems I have missed some basics of factorials. I was surprised to find out that (-1)!=-1. I was thinking it is undefined.
If (-1)!=-1 is it correct to assume the following:
(-2)! = 2 = -1 X -2
(-3)! = -6 = -1 X -2 X -3
(-4)! = 24 = -1 X -2 X -3 X -4
(-5)! = -120 = -1 X -2 X -3 X -4 X -5
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55732
Re: D01-16  [#permalink]

Show Tags

New post 11 Dec 2014, 10:23
1
Van4ez wrote:
It seems I have missed some basics of factorials. I was surprised to find out that (-1)!=-1. I was thinking it is undefined.
If (-1)!=-1 is it correct to assume the following:
(-2)! = 2 = -1 X -2
(-3)! = -6 = -1 X -2 X -3
(-4)! = 24 = -1 X -2 X -3 X -4
(-5)! = -120 = -1 X -2 X -3 X -4 X -5


No, that;s not correct at all. Factorial is defined only for non-negative integers: the factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n.

In the original question we don't have (-1)! (which would be undefined), we have (|-1|)!. Notice that -1 there is in modulus sign: (|-1|)! = (1)! = 1.

Hope it's clear.
_________________
Intern
Intern
avatar
Joined: 27 Oct 2014
Posts: 23
GMAT 1: 680 Q49 V34
GPA: 3.8
GMAT ToolKit User Reviews Badge
Re: D01-16  [#permalink]

Show Tags

New post 11 Dec 2014, 10:29
Thank you, Bunuel, my fault. This is a good example why I couldn't reach my desired score:)
Manager
Manager
User avatar
S
Joined: 10 Sep 2015
Posts: 73
Location: India
Concentration: Finance, Human Resources
GMAT 1: 640 Q47 V31
GMAT 2: 660 Q47 V35
GMAT 3: 700 Q49 V36
GPA: 4
Reviews Badge
Re: D01-16  [#permalink]

Show Tags

New post 29 Nov 2017, 15:49
Bunuel wrote:
Official Solution:


If \((|p|!)^p = |p|!\), which of the following could be true?

I. \(p=-1\)

II. \(p=0\)

III. \(p=1\)


A. I only
B. II only
C. III only
D. II and III only
E. I, II and III


Important properties: \(0!=1\) and any non-zero number to the power of 0 is 1.

Let's check the options:

If \(p=-1\) then \((|p|!)^p = (|-1|!)^{-1}=1^{-1}=1\) and \(|p|!=|-1|!=1!=1\) so \(p\) could be -1;

If \(p=0\) then \((|p|!)^p = (|0|!)^{0}=1^{0}=1\) and \(|0|!=0!=1\) so \(p\) could be 0;

If \(p=1\) then \((|p|!)^p = (|1|!)^{1}=1^{1}=1\) and \(|p|!=|1|!=1\) so \(p\) could be 1.


Answer: E


Hi
In one of your article you mentioned that factorial of negative numbers is undefined so how here you are considering-1!=1?
>> !!!

You do not have the required permissions to view the files attached to this post.

Manager
Manager
User avatar
S
Joined: 10 Sep 2015
Posts: 73
Location: India
Concentration: Finance, Human Resources
GMAT 1: 640 Q47 V31
GMAT 2: 660 Q47 V35
GMAT 3: 700 Q49 V36
GPA: 4
Reviews Badge
Re: D01-16  [#permalink]

Show Tags

New post 29 Nov 2017, 15:50
Bunuel wrote:
Official Solution:


If \((|p|!)^p = |p|!\), which of the following could be true?

I. \(p=-1\)

II. \(p=0\)

III. \(p=1\)


A. I only
B. II only
C. III only
D. II and III only
E. I, II and III


Important properties: \(0!=1\) and any non-zero number to the power of 0 is 1.

Let's check the options:

If \(p=-1\) then \((|p|!)^p = (|-1|!)^{-1}=1^{-1}=1\) and \(|p|!=|-1|!=1!=1\) so \(p\) could be -1;

If \(p=0\) then \((|p|!)^p = (|0|!)^{0}=1^{0}=1\) and \(|0|!=0!=1\) so \(p\) could be 0;

If \(p=1\) then \((|p|!)^p = (|1|!)^{1}=1^{1}=1\) and \(|p|!=|1|!=1\) so \(p\) could be 1.


Answer: E


Hi
In one of your article you mentioned that factorial of negative numbers is undefined so how here you are considering-1!=1?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55732
Re: D01-16  [#permalink]

Show Tags

New post 29 Nov 2017, 20:45
asthagupta wrote:
Bunuel wrote:
Official Solution:


If \((|p|!)^p = |p|!\), which of the following could be true?

I. \(p=-1\)

II. \(p=0\)

III. \(p=1\)


A. I only
B. II only
C. III only
D. II and III only
E. I, II and III


Important properties: \(0!=1\) and any non-zero number to the power of 0 is 1.

Let's check the options:

If \(p=-1\) then \((|p|!)^p = (|-1|!)^{-1}=1^{-1}=1\) and \(|p|!=|-1|!=1!=1\) so \(p\) could be -1;

If \(p=0\) then \((|p|!)^p = (|0|!)^{0}=1^{0}=1\) and \(|0|!=0!=1\) so \(p\) could be 0;

If \(p=1\) then \((|p|!)^p = (|1|!)^{1}=1^{1}=1\) and \(|p|!=|1|!=1\) so \(p\) could be 1.


Answer: E


Hi
In one of your article you mentioned that factorial of negative numbers is undefined so how here you are considering-1!=1?


Explained here: https://gmatclub.com/forum/d01-183477.html#p1454619
_________________
Intern
Intern
avatar
Joined: 17 Nov 2017
Posts: 1
Re: D01-16  [#permalink]

Show Tags

New post 29 Nov 2017, 21:20
asthagupta wrote:
Bunuel wrote:
Official Solution:


If \((|p|!)^p = |p|!\), which of the following could be true?

I. \(p=-1\)

II. \(p=0\)

III. \(p=1\)


A. I only
B. II only
C. III only
D. II and III only
E. I, II and III


Important properties: \(0!=1\) and any non-zero number to the power of 0 is 1.

Let's check the options:

If \(p=-1\) then \((|p|!)^p = (|-1|!)^{-1}=1^{-1}=1\) and \(|p|!=|-1|!=1!=1\) so \(p\) could be -1;

If \(p=0\) then \((|p|!)^p = (|0|!)^{0}=1^{0}=1\) and \(|0|!=0!=1\) so \(p\) could be 0;

If \(p=1\) then \((|p|!)^p = (|1|!)^{1}=1^{1}=1\) and \(|p|!=|1|!=1\) so \(p\) could be 1.


Answer: E


Hi
In one of your article you mentioned that factorial of negative numbers is undefined so how here you are considering-1!=1?

Mode of any no. i.e. |-1|= 1.

Sent from my SM-G925I using GMAT Club Forum mobile app
Manager
Manager
User avatar
S
Joined: 10 Sep 2015
Posts: 73
Location: India
Concentration: Finance, Human Resources
GMAT 1: 640 Q47 V31
GMAT 2: 660 Q47 V35
GMAT 3: 700 Q49 V36
GPA: 4
Reviews Badge
Re: D01-16  [#permalink]

Show Tags

New post 30 Nov 2017, 12:02
i did not notice the modulus sign :(

Got it now, thanks Bunuel for quick turnaround
Manager
Manager
avatar
S
Joined: 29 Sep 2017
Posts: 117
Location: United States
GMAT ToolKit User Reviews Badge
Re: D01-16  [#permalink]

Show Tags

New post 08 Jun 2018, 15:31
The interesting thing about this question is that you can forget that 0! = 1 and still get it right. It's easy to prove that 1 and 3 are correct and given that none of the answer choices are 1 and 3 only, it must be E.
_________________
If this helped, please give kudos!
Intern
Intern
avatar
B
Joined: 26 Nov 2017
Posts: 4
Re D01-16  [#permalink]

Show Tags

New post 21 Sep 2018, 07:28
I think this is a high-quality question and I agree with explanation. Best question tests the concept of absolute value, exponents and factorials
GMAT Club Bot
Re D01-16   [#permalink] 21 Sep 2018, 07:28
Display posts from previous: Sort by

D01-16

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

Moderators: chetan2u, Bunuel



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