How perfect are the reciprocals of the factors??

If n n is a perfect number and d 1 , d 2 , . . . . , d k d_1,d_2,....,d_k denote all the divisors of n n including n n itself,find the value of 1 d 1 + 1 d 2 + . . . . . . . . . . . + 1 d k . \frac{1}{d_1} + \frac{1}{d_2} + ........... + \frac{1}{d_k}.


The answer is 2.

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2 solutions

Josh Rowley
Apr 18, 2014

We know that the divisors come in pairs, ie. d i × d j = n d_{i} \times d_{j} = n for some i , j i,j . Therefore, WLOG, say d 2 i 1 × d 2 i = n d_{2i-1} \times d_{2i} = n for all 1 i k 2 1 \le i \le \dfrac{k}{2} . Then we can combine the fractions of each pair, meaning the statement becomes: d 1 + d 2 d 1 × d 2 + d 3 + d 4 d 3 × d 4 + . . . + d k 1 + d k d k 1 × d k \dfrac{d_{1}+d_{2}}{d_{1} \times d_{2}} + \dfrac{d_{3}+d_{4}}{d_{3} \times d_{4}} + ... + \dfrac{d_{k-1}+d_{k}}{d_{k-1} \times d_{k}} But the denominator of all these fractions is just n n , so we can group all the fractions, meaning the statement becomes: i = 1 k d i n \displaystyle\sum_{i=1}^{k} \dfrac{d_{i}}{n} We know that n n is a perfect number, so i = 1 k = 2 n \displaystyle\sum_{i=1}^{k} = 2n . Therefore the statement becomes: 2 n n = 2 \dfrac{2n}{n} = \fbox{2}

You could have done it way more easily.As n is perfect, then d 1 + d 2 . . . + d k = 2 n . d_1+d_2...+d_k=2n. .Divide both sides by n and you get the answer.

Bogdan Simeonov - 7 years, 1 month ago

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How?please explain

Adarsh Kumar - 7 years, 1 month ago
Geo Cheruvathoor
Apr 21, 2014

take an example. for 6, divisors are 1,2,3 and 6. so 1/1+1/2+1/3+1/6=2

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