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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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Updated on: 07 Oct 2018, 05:48
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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
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06 Oct 2018, 10:23
Official Explanation In general, perfect number is (2^(n1))(2^n1) when n=prime number. In other words, when n=2, (2^(21))(2^21)=(2)(3)=6 is the first perfect number. The 2nd perfect number is when n=3, and with substitution, you get (2^(31))(2^31)=(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
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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
Hence B is the answer
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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Re: If the sum of all positive factors of an integer n is 2n, n is a perfe
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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
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Re: If the sum of all positive factors of an integer n is 2n, n is a perfe
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06 Oct 2018, 10:40
E because factors of 12 are 12, 6, 3, 2, and 1. \(12+6+3+2+1=24=12*2\)
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Re: If the sum of all positive factors of an integer n is 2n, n is a perfe
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06 Oct 2018, 10:42
Is the answer d? 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
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06 Oct 2018, 10:46
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
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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



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Re: If the sum of all positive factors of an integer n is 2n, n is a perfe
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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
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Re: If the sum of all positive factors of an integer n is 2n, n is a perfe
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30 Jan 2019, 11:06
gmatbusters wrote: Official Explanation In general, perfect number is (2^(n1))(2^n1) when n=prime number. In other words, when n=2, (2^(21))(2^21)=(2)(3)=6 is the first perfect number. The 2nd perfect number is when n=3, and with substitution, you get (2^(31))(2^31)=(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^(n1))(2^n1)?



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Re: If the sum of all positive factors of an integer n is 2n, n is a perfe
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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^(n1))(2^n1) when n=prime number. In other words, when n=2, (2^(21))(2^21)=(2)(3)=6 is the first perfect number. The 2nd perfect number is when n=3, and with substitution, you get (2^(31))(2^31)=(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^(n1))(2^n1)? Posted from my mobile device
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Re: If the sum of all positive factors of an integer n is 2n, n is a perfe
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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
Hence B is the answer 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
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28 Mar 2019, 00:57
chetan2u VeritasKarishma mikemcgarry gmatbusters wrote: Official Explanation In general, perfect number is (2^(n1))(2^n1) when n=prime number. In other words, when n=2, (2^(21))(2^21)=(2)(3)=6 is the first perfect number. The 2nd perfect number is when n=3, and with substitution, you get (2^(31))(2^31)=(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
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28 Mar 2019, 04:24
sakshamchhabra wrote: chetan2u VeritasKarishma mikemcgarry gmatbusters wrote: Official Explanation In general, perfect number is (2^(n1))(2^n1) when n=prime number. In other words, when n=2, (2^(21))(2^21)=(2)(3)=6 is the first perfect number. The 2nd perfect number is when n=3, and with substitution, you get (2^(31))(2^31)=(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.htmlIn 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
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28 Mar 2019, 07:20
sakshamchhabra wrote: chetan2u VeritasKarishma mikemcgarry gmatbusters wrote: Official Explanation In general, perfect number is (2^(n1))(2^n1) when n=prime number. In other words, when n=2, (2^(21))(2^21)=(2)(3)=6 is the first perfect number. The 2nd perfect number is when n=3, and with substitution, you get (2^(31))(2^31)=(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 \((2^{n1})(2^n1)\), what will be the value when n is 3 or so.
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Re: If the sum of all positive factors of an integer n is 2n, n is a perfe
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