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# D01-16

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Math Expert
Joined: 02 Sep 2009
Posts: 50711

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15 Sep 2014, 23:12
00:00

Difficulty:

25% (medium)

Question Stats:

68% (00:56) correct 32% (00:32) wrong based on 157 sessions

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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

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Joined: 02 Sep 2009
Posts: 50711

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15 Sep 2014, 23:12
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.

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11 Dec 2014, 05:11
1^-1 = 1 ? Please explain.
Math Expert
Joined: 02 Sep 2009
Posts: 50711

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11 Dec 2014, 05:30
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

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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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11 Dec 2014, 09:23
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.
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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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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.

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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Joined: 10 Sep 2015
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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.

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
Joined: 02 Sep 2009
Posts: 50711

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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.

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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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.

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

Got it now, thanks Bunuel for quick turnaround
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08 Jun 2018, 14:31
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.
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21 Sep 2018, 06:28
I think this is a high-quality question and I agree with explanation. Best question tests the concept of absolute value, exponents and factorials
Re D01-16 &nbs [#permalink] 21 Sep 2018, 06:28
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# D01-16

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