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

Question

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

Show Answer

Official Answer

C

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

Weekly Quant Quiz #3 Question No 5

GMATBusters · Accepted Answer

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.

pk123 · Answer

12 factors = 1,2, 3, 6, 12
sp 1+2+3+6+12=24 which is 2*12

Hence B is the answer

Gladiator59 · Answer

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

MsInvBanker · Answer

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

12+6+3+2+1=24=12*2

pineapple123456 · Answer

Is the answer d?
Cheers

Using the options.

rever08 · Answer

28 is the other such number, factors of 28 are 6. Answer C

hargun3045 · Answer

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

u1983 · Answer

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

arpitkansal · Answer

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

GMATBusters · Answer

This is just another approach.

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

Posted from my mobile device

sakshamchhabra · Answer

you forgot to consider 24, a factor of 24

which makes 24 not a perfect number

sakshamchhabra · Answer

chetan2u  
 VeritasKarishma   mikemcgarry

please throw some light on how to derive this formula

KarishmaB · Answer

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.

chetan2u · Answer

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.

Paras10 · Answer

You missed out the number 4 ...

Factors of 12 are 1, 2 , 3 , 4 , 6 , 12

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