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Retired Moderator V
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If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

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Question Stats: 47% (02:24) correct 53% (02:35) wrong based on 86 sessions

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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

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

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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.
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Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

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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.
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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]

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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.

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Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

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E because factors of 12 are 12, 6, 3, 2, and 1.

$$12+6+3+2+1=24=12*2$$
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Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

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Cheers

Using the options.
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Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

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28 is the other such number, factors of 28 are 6. Answer C
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Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

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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
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GMAT 1: 540 Q49 V16 GMAT 2: 680 Q49 V33 Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

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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

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Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

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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)?
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Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

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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)?

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Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

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

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

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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.
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Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

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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
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