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

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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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Updated on: 07 Oct 2018, 04:48
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Question Stats:

45% (02:08) correct 55% (02:24) wrong based on 64 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, 09:21.
Last edited by gmatbusters on 07 Oct 2018, 04:48, 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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06 Oct 2018, 09:23

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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Updated on: 06 Oct 2018, 11:31
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, 09:26.
Last edited by pk123 on 06 Oct 2018, 11:31, edited 1 time in total.
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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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06 Oct 2018, 09:31
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.

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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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06 Oct 2018, 09:40
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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06 Oct 2018, 09:42
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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06 Oct 2018, 09:46
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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06 Oct 2018, 10:48
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

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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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06 Oct 2018, 11:14
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

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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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30 Jan 2019, 10:06
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)?
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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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30 Jan 2019, 19:17
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)?

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Re: If the sum of all positive factors of an integer n is 2n, n is a perfe   [#permalink] 30 Jan 2019, 19:17
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