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Retired Moderator
Joined: 16 Apr 2020
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Sum of all positive factors of N VS 8232
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01 Sep 2021, 09:35
01 Sep 2021, 09:35
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Question Stats:
28% (02:32) correct
71% (02:13) wrong based on 85 sessions
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\(N = 3^8 - 2^8\)
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Sum of all positive factors of N |
8232 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Sum of all positive factors of N VS 8232
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06 Sep 2021, 07:27
06 Sep 2021, 07:27
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grenico wrote:
We can use the difference of squares to solve this easily.
Note that for any integer \(n\), we have:
\((a^{n} - b^{n}) = (a^{n/2} - b^{n/2})(a^{n/2} + b^{n/2})\)
So we have:
\((3^{8} - 2^{8})\)
\((3^{4} - 2^{4})(3^{4} + 2^{4})\)
\((3^{2} - 2^{2})(3^{2} + 2^{2})(3^{4} + 2^{4})\)
\((3 - 2)(3 + 2)(3^{2} + 2^{2})(3^{4} + 2^{4})\)
\((1)(5)(9+4)(81+16)\)
\((1)(5)(13)(97)\)
Since 97 is prime, we can no longer factor this expression.
Now we need to multiply the different combinations of these factors to get the complete list of factors for \((3^{8} - 2^{8})\).
Think of factoring 12. This would be \(2^{2} * 3\), or \(4*3\). Summing 4 and 3 isn't enough, we're forgetting the 1,2,6 and 12.
Permuting the different factors then, we get:
\(1,5,13,97,5*13,5*97, 13*97, 5*13*97\)
Which simplifies too (I used a calculator here, definitely saves time):
\(1,5,13,97,65, 485, 1261, 6305\)
Summing these we get \(8232\).
Therefore the answer is C
Note that for any integer \(n\), we have:
\((a^{n} - b^{n}) = (a^{n/2} - b^{n/2})(a^{n/2} + b^{n/2})\)
So we have:
\((3^{8} - 2^{8})\)
\((3^{4} - 2^{4})(3^{4} + 2^{4})\)
\((3^{2} - 2^{2})(3^{2} + 2^{2})(3^{4} + 2^{4})\)
\((3 - 2)(3 + 2)(3^{2} + 2^{2})(3^{4} + 2^{4})\)
\((1)(5)(9+4)(81+16)\)
\((1)(5)(13)(97)\)
Since 97 is prime, we can no longer factor this expression.
Now we need to multiply the different combinations of these factors to get the complete list of factors for \((3^{8} - 2^{8})\).
Think of factoring 12. This would be \(2^{2} * 3\), or \(4*3\). Summing 4 and 3 isn't enough, we're forgetting the 1,2,6 and 12.
Permuting the different factors then, we get:
\(1,5,13,97,5*13,5*97, 13*97, 5*13*97\)
Which simplifies too (I used a calculator here, definitely saves time):
\(1,5,13,97,65, 485, 1261, 6305\)
Summing these we get \(8232\).
Therefore the answer is C
In order to find the sum of factors of any number \(N\), where
\(N = a^xb^yc^z.....\) and \(a, b, c\) are prime factors
Sum of (+ve) factors = \((\frac{a^{x+1} - 1}{a - 1})(\frac{b^{y+1} - 1}{b - 1})(\frac{c^{z+1} - 1}{c - 1})..\)
Here, \(N = (5)(13)(97)\)
Sum = \((\frac{5^2 - 1}{5 - 1})(\frac{13^2 - 1}{13 - 1})(\frac{97^2 - 1}{97 - 1})\)
= \(\frac{(24)(168)(9408)}{(4)(12)(96)} = 8232\)
General Discussion
Re: Sum of all positive factors of N VS 8232
[#permalink]
01 Sep 2021, 10:05
01 Sep 2021, 10:05
answer is C, but I am not sure what's the logic.
I could do only with calculator
I could do only with calculator
