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D01-16 [#permalink]

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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
[Reveal] Spoiler: OA

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New post 15 Sep 2014, 23:12
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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
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Re: D01-16 [#permalink]

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New post 11 Dec 2014, 05:11
1^-1 = 1 ? Please explain.

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New post 11 Dec 2014, 05:30
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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

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Re: D01-16 [#permalink]

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New post 11 Dec 2014, 08:00
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

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New post 11 Dec 2014, 09:23
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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.
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Re: D01-16 [#permalink]

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New post 11 Dec 2014, 09:29
Thank you, Bunuel, my fault. This is a good example why I couldn't reach my desired score:)

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Re: D01-16 [#permalink]

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New post 29 Nov 2017, 14:49
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?
>> !!!

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Re: D01-16 [#permalink]

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New post 29 Nov 2017, 14:50
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?

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Re: D01-16 [#permalink]

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New post 29 Nov 2017, 19:45
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
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: D01-16 [#permalink]

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New post 29 Nov 2017, 20:20
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.

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Re: D01-16 [#permalink]

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New post 30 Nov 2017, 11:02
i did not notice the modulus sign :(

Got it now, thanks Bunuel for quick turnaround

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Re: D01-16   [#permalink] 30 Nov 2017, 11:02
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