Example: 28 = 1 + 2 + 4 + 7 + 14 ==> if one fully paid equal to the sum of its divisor is my c-code this: Code: # Include <stdio.h> # Include <conio.h> int main () ( long z, i, sum = 0; for(z = 1, z <= 1000000; z + +) ( for(i = 1; i <z; i + +) ( if(z% i == 0) sum + = i; ) if(sum == x) printf ("% \ n ld", z); sum = 0; ) return 0; ) now I have the problem that it takes very long (on my athlon 1800 + I got after 5 min terminated because he was only at about 80 000) can this be done efficiently?

Code: for (i = 1; i <z; i + +) if (z% i == 0) sum + = i; looks like for Maich Code: for (i = 1, i * i <z; i + +) if (z% i == 0) (sum + = i; sum + = z / i;) if (z% i == 0) sum + = i;

Code: for (z = 1, z <= 1000000; z + +) ( for (i = 1; i <z; i + +) ( if (z% i == 0) sum + = i; ) if (sum == x) printf ("% ld \ n", z); sum = 0; ) You could be in the "i loop" Check whether the sum is too large and, if yes cancel. Code: for (i = 1; i <z; i + +) ( if (z% i == 0) (Sum + = i; if (sum> z) = z i / / or break, (I do not break :-)) ) ) I too have to believe that the i-loop search only to z / 2: for (i = 1, (i * 2) <= z; i + +) and I would look for from big to small for (i = (z +1) / 2; i> 0, - i) / / the z +1 to be sure, / / is that even the half of tested But then I must terminate with i = 0 to change

Code: # Include <stdio.h> # Include <math.h> int main () ( long z, i, sum = 1, / / there is always a divisor for (z = 1, z <= 1000000; z + +) ( for (i = sqrt (z), i> 1, - i) / / without a splitter to the root (If (z% i == 0) (Sum + = i; sum + = z / i / / splitter, each has its "co-parter" if (sum> z) i = 0; / / demolition as the sum too large ) ) if (sum == x) printf ("% ld \ n", z); sum = 1, / / there is always a divisor ) return 0; ) NEN on Pentium 450s in about two minutes the whole Mille calculated through but only when the figures found if I do not know all of the

That's all. But since you'll have to come up, many time-saving things, so you can find as many. The fifth is 33,550,336, and already the sixth 8589869056 ...

It may interest one or the other too: Perfect numbers are always a sum of consecutive numbers, eg: 6 = 1 +2 +3, 28 = 1 +2 +3 +4 +5 +6 +7, 496 = 1 +2 +3 +4 +5 +...+ 30 +31 8128 = 1 +2 +3 +4 +5 +...+ 126 +127 Euclid proved using the geometric series: n for some numbers p = 1 +2 +4 +8 +...+ 2 ^ n = 2 ^ (n +1) -1 is a prime number. In each such case, 2 ^ n Â· p is perfect. 6 = 2 * (2 Â² -1) 28 = 2 Â² * (2 Â³ -1) It was later proved by Euler, that this rule all even, perfect numbers provides - if you find a suitable prime. Therefore, virtually every new large prime Fund is associated with a further big perfect number. The question of whether it is also odd perfect numbers, is still unclear. In case of existence of these would need to be larger than 10 ^ 100 and have at least 11 different prime divisors.

Yes. Top I add the paired dividers. So for 36 I'll do as 2 18 3 12 4 9 and down again I'll do the 6 and the one I've forgotten. yes, if every perfect number ne triangular number must be, because then you have still NEN fine accelerator.

f I want to study up to z / 2 why I did then as a termination condition not> or <z / 2 z and square root but i?

because sqrt (z) is already sufficient. z / 2 would be a waste of time. dividers for each i with z% i == 0, the opposites meet g = z and z% i% g == 0. so you only need to go to root, as would i == g. Otherwise, is always a small and a larger than the root.

creative, Make sure you understand the difference between CODE & QUOTE tag. Refer http://www.cfanatic.com/misc.php?do=bbcode if you have any doubts.