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
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 = \((\frac{5^2 - 1}{5 - 1})(\frac{13^2 - 1}{13 - 1})(\frac{97^2 - 1}{97 - 1})\)
= \(\frac{(24)(168)(9408)}{(4)(12)(96)} = 8232\)