If n is a perfect number and d 1 , d 2 , . . . . , d k denote all the divisors of n including n itself,find the value of d 1 1 + d 2 1 + . . . . . . . . . . . + d k 1 .
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You could have done it way more easily.As n is perfect, then d 1 + d 2 . . . + d k = 2 n . .Divide both sides by n and you get the answer.
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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We know that the divisors come in pairs, ie. d i × d j = n for some i , j . Therefore, WLOG, say d 2 i − 1 × d 2 i = n for all 1 ≤ i ≤ 2 k . 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 But the denominator of all these fractions is just n , so we can group all the fractions, meaning the statement becomes: i = 1 ∑ k n d i We know that n is a perfect number, so i = 1 ∑ k = 2 n . Therefore the statement becomes: n 2 n = 2