How to write an efficient script for perfect numbers

I have to display the first 4 perfect numbers but am running into a problem. The way I am trying it takes very very long to reach the last value (8128 or whatever). Figured out how to post code (there's a freaking button for it lol...)

for k = 1:9000
  factors = divisors(k);
  factors(end) = [];
  if sum(factors) == k
      disp(k)
end

[KSSV]

You have to show code to figure out your difficulty..

[K M]

As I was about to comment I just saw the button for 'code' lol... Anyway here's what I have:

 for k = 1:9000
    factors = divisors(k);
    factors(end) = [];
    if sum(factors) == k
        disp(k)
  end

Answer 1

[KSSV]

Try this:

for k = 1:9000
    % factors = divisors(k);
    D = 1:k ;
    factors =D(rem(k,D)==0) ;
    factors(end) = [] ;
    if sum(factors) == k
        disp(k)
    end
end

[K M]

yes! that was instantaneous! however it may take me some time to understand the logic... thank you for your help

[Grace Lim]

@KSSV, sorry to bother you but can you explain this line '

factors =D(rem(k,D)==0)

[Walter Roberson]

rem(k,D) is the remainder when k is divided by D, which will be 0 when D exactly divides k. This code uses logical indexing to extract the exact divisors.