Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 18 Jul 2019, 00:00

### 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

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

# If the sum of all positive factors of an integer n is 2n, n is a perfe

Author Message
TAGS:

### Hide Tags

Retired Moderator
Joined: 27 Oct 2017
Posts: 1229
Location: India
GPA: 3.64
WE: Business Development (Energy and Utilities)
If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

Updated on: 07 Oct 2018, 05:48
4
00:00

Difficulty:

85% (hard)

Question Stats:

44% (02:20) correct 56% (02:32) wrong based on 79 sessions

### HideShow timer Statistics

If the sum of all positive factors of an integer n is 2n, n is a perfect number. For example, the factors of 6 are 1, 2, 3, and 6, and from the sum 1+2+3+6=12=2*6, the sum of the factors of 6 becomes 12=2*6, thus 6 is the first perfect number. Then, what is the number of factors of the second perfect number?

A. 4
B. 5
C. 6
D. 8
E. 12

Weekly Quant Quiz #3 Question No 5

_________________

Originally posted by gmatbusters on 06 Oct 2018, 10:21.
Last edited by gmatbusters on 07 Oct 2018, 05:48, edited 2 times in total.
Renamed the topic and edited the question.
Retired Moderator
Joined: 27 Oct 2017
Posts: 1229
Location: India
GPA: 3.64
WE: Business Development (Energy and Utilities)
If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

06 Oct 2018, 10:23
1
1

Official Explanation

In general, perfect number is (2^(n-1))(2^n-1) when n=prime number.
In other words, when n=2, (2^(2-1))(2^2-1)=(2)(3)=6 is the first perfect number.
The 2nd perfect number is when n=3, and with substitution, you get (2^(3-1))(2^3-1)=(4)(7)=28.
In other words, the sum of all factors of 28=1+2+4+7+14+28=2(28), hence it is a perfect number.
If $$N = a^p*b^q...$$, No of factors =$$(p+1)(q+1)...$$
From 28 = $$(2^2)(7^1)$$, the number of factors is (2+1)(1+1)=6. The answer is C.
_________________
Manager
Joined: 16 Sep 2011
Posts: 91
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

Updated on: 06 Oct 2018, 12:31
12 factors = 1,2, 3, 6, 12
sp 1+2+3+6+12=24 which is 2*12

Originally posted by pk123 on 06 Oct 2018, 10:26.
Last edited by pk123 on 06 Oct 2018, 12:31, edited 1 time in total.
Senior PS Moderator
Status: It always seems impossible until it's done.
Joined: 16 Sep 2016
Posts: 751
GMAT 1: 740 Q50 V40
GMAT 2: 770 Q51 V42
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

06 Oct 2018, 10:31
Second perfect number comes out to be 28 after a little trial and error. The factors of 28 are 1,2,4,7,14,28 and hence it has 6 factors.

So Option (C) is correct.

Best,
G

Posted from my mobile device
_________________
Regards,

“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back)
Retired Moderator
Joined: 23 May 2018
Posts: 487
Location: Pakistan
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

06 Oct 2018, 10:40
E because factors of 12 are 12, 6, 3, 2, and 1.

$$12+6+3+2+1=24=12*2$$
_________________
If you can dream it, you can do it.

Practice makes you perfect.

Kudos are appreciated.
Manager
Joined: 26 Sep 2018
Posts: 60
Location: Sweden
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

06 Oct 2018, 10:42
Cheers

Using the options.
Manager
Joined: 21 Jul 2017
Posts: 190
Location: India
GMAT 1: 660 Q47 V34
GPA: 4
WE: Project Management (Education)
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

06 Oct 2018, 10:46
28 is the other such number, factors of 28 are 6. Answer C
Intern
Joined: 06 Feb 2018
Posts: 16
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

06 Oct 2018, 11:48
6 2*3 = 1 + 2 + 3

The key here is that 2+1 = 3

A few trials with other primes shows 5 = 2 + 2 + 1
but we can't have two repeat has a factor
next option is 4
Therefore 1 + 2 + 4 = 7
i.e 2^2*7 = 1 + 2 + 4 + 7 + 14
number is 28
Factors 3*2 = 6

C
RC Moderator
Joined: 24 Aug 2016
Posts: 802
Concentration: Entrepreneurship, Operations
GMAT 1: 630 Q48 V28
GMAT 2: 540 Q49 V16
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

06 Oct 2018, 12:14
Really not sure if there is a mathematical way of deduction.......
..................but by iteration got the next perfect number as 28.
Factors of 28= 1,2,4,7,14,28
There are 6 factors
Ans C

_________________
Please let me know if I am going in wrong direction.
Thanks in appreciation.
Manager
Joined: 17 Jun 2018
Posts: 56
Location: India
Schools: IMD '20
GPA: 2.84
WE: Engineering (Consulting)
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

30 Jan 2019, 11:06
gmatbusters wrote:

Official Explanation

In general, perfect number is (2^(n-1))(2^n-1) when n=prime number.
In other words, when n=2, (2^(2-1))(2^2-1)=(2)(3)=6 is the first perfect number.
The 2nd perfect number is when n=3, and with substitution, you get (2^(3-1))(2^3-1)=(4)(7)=28.
In other words, the sum of all factors of 28=1+2+4+7+14+28=2(28), hence it is a perfect number.
If $$N = a^p*b^q...$$, No of factors =$$(p+1)(q+1)...$$
From 28 = $$(2^2)(7^1)$$, the number of factors is (2+1)(1+1)=6. The answer is C.

how did you arrive at the formulation (2^(n-1))(2^n-1)?
Retired Moderator
Joined: 27 Oct 2017
Posts: 1229
Location: India
GPA: 3.64
WE: Business Development (Energy and Utilities)
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

30 Jan 2019, 20:17
This is just another approach.

We can solve the question by iteration as shown by gladiator.

arpitkansal wrote:
gmatbusters wrote:

Official Explanation

In general, perfect number is (2^(n-1))(2^n-1) when n=prime number.
In other words, when n=2, (2^(2-1))(2^2-1)=(2)(3)=6 is the first perfect number.
The 2nd perfect number is when n=3, and with substitution, you get (2^(3-1))(2^3-1)=(4)(7)=28.
In other words, the sum of all factors of 28=1+2+4+7+14+28=2(28), hence it is a perfect number.
If $$N = a^p*b^q...$$, No of factors =$$(p+1)(q+1)...$$
From 28 = $$(2^2)(7^1)$$, the number of factors is (2+1)(1+1)=6. The answer is C.

how did you arrive at the formulation (2^(n-1))(2^n-1)?

Posted from my mobile device
_________________
Manager
Joined: 09 Mar 2018
Posts: 56
Location: India
Schools: CBS Deferred "24
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

28 Mar 2019, 00:50
pk123 wrote:
12 factors = 1,2, 3, 6, 12
sp 1+2+3+6+12=24 which is 2*12

you forgot to consider 24, a factor of 24

which makes 24 not a perfect number
Manager
Joined: 09 Mar 2018
Posts: 56
Location: India
Schools: CBS Deferred "24
If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

28 Mar 2019, 00:57
chetan2u
gmatbusters wrote:

Official Explanation

In general, perfect number is (2^(n-1))(2^n-1) when n=prime number.
In other words, when n=2, (2^(2-1))(2^2-1)=(2)(3)=6 is the first perfect number.
The 2nd perfect number is when n=3, and with substitution, you get (2^(3-1))(2^3-1)=(4)(7)=28.
In other words, the sum of all factors of 28=1+2+4+7+14+28=2(28), hence it is a perfect number.
If $$N = a^p*b^q...$$, No of factors =$$(p+1)(q+1)...$$
From 28 = $$(2^2)(7^1)$$, the number of factors is (2+1)(1+1)=6. The answer is C.

please throw some light on how to derive this formula
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9442
Location: Pune, India
If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

28 Mar 2019, 04:24
1
sakshamchhabra wrote:
chetan2u
gmatbusters wrote:

Official Explanation

In general, perfect number is (2^(n-1))(2^n-1) when n=prime number.
In other words, when n=2, (2^(2-1))(2^2-1)=(2)(3)=6 is the first perfect number.
The 2nd perfect number is when n=3, and with substitution, you get (2^(3-1))(2^3-1)=(4)(7)=28.
In other words, the sum of all factors of 28=1+2+4+7+14+28=2(28), hence it is a perfect number.
If $$N = a^p*b^q...$$, No of factors =$$(p+1)(q+1)...$$
From 28 = $$(2^2)(7^1)$$, the number of factors is (2+1)(1+1)=6. The answer is C.

please throw some light on how to derive this formula

The formula is not entirely accurate. It does not hold for all prime values of n such as it will not hold for n = 11.

Its logic lies in the fact that all powers of 2 add up to give 1 less than the next power of 2 (which is something you should know, I agree).

e.g. 2^0 + 2^1 + 2^2 = 2^3 - 1
2^0 + 2^1 + 2^2 + 2^3 = 2^4 - 1
and so on...

So when the right hand side is prime, the sum of all factors is the number itself. Try it out with n = 5 to see how.
This does not happen for all prime values of n though.

See here for more on perfect numbers: https://www.mathsisfun.com/numbers/prim ... anced.html

In my opinion, it is not a fair question for GMAT.
_________________
Karishma
Veritas Prep GMAT Instructor

Math Expert
Joined: 02 Aug 2009
Posts: 7764
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

28 Mar 2019, 07:20
1
sakshamchhabra wrote:
chetan2u
gmatbusters wrote:

Official Explanation

In general, perfect number is (2^(n-1))(2^n-1) when n=prime number.
In other words, when n=2, (2^(2-1))(2^2-1)=(2)(3)=6 is the first perfect number.
The 2nd perfect number is when n=3, and with substitution, you get (2^(3-1))(2^3-1)=(4)(7)=28.
In other words, the sum of all factors of 28=1+2+4+7+14+28=2(28), hence it is a perfect number.
If $$N = a^p*b^q...$$, No of factors =$$(p+1)(q+1)...$$
From 28 = $$(2^2)(7^1)$$, the number of factors is (2+1)(1+1)=6. The answer is C.

please throw some light on how to derive this formula

As karishma too has pointed out, nothing that you should worry about. What max you should know is that the first two perfect numbers are 6 and 28, and their factors less itself would sum up to the number itself.
The above information is good enough but still not likely to be tested on the GMAT, and you do not require to know any further.

If ever GMAT wants to test this, it, at the most, would give you this as a function.
The question could be - If a perfect number is defined by the function \$ and is given by $$(2^{n-1})(2^n-1)$$, what will be the value when n is 3 or so.
_________________
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe   [#permalink] 28 Mar 2019, 07:20
Display posts from previous: Sort by