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Need Help on a Tedious Problem (1 Viewer)

HAX0R

Member
Joined
Sep 24, 2013
Messages
125
Gender
Male
HSC
2018
Hello

Got another set of questions and this question is really bugging me.

An amicable pair (x,y) is where the proper divisiors (divisors excluding itself) of y sum up to x, and the proper divisors of x sum up to y. Given that p=3*2^(n)-1 and q=3*2^(n-1)-1 and r=9*2^(2n-1)-1 are ALL prime, prove that x=2^(n)pq and 2^(n)r are amicable.

I proved that the sum of the proper divisors of y is always equal to x. I wrote down all the factors of 2^(n)r, which are 2, 2 squared, 2 cubed... 2^n, r, 1, 2r, 2 squared r, 2 cubed r.. 2^(n-1)r. Then I used geometric sequence to sum it all up, and then I went on wolfram to calculate it all up. Then I calculated 2^(n)pq, and it matched.

However, when ig attempted it for 2^(n)pq, it didn't work out. I'm feeling I missed a factor or I did something wrong. Would anyone bother to show me how it's done? (Just the proof that the proper divisors of 2^(n)pq sum to 2^(n)r.

Thanks.
 

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