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factoring

vaneger · 2004-04-03T22:01:20

how can i get the factors of a number?

For Beginners

Started by vaneger April 03, 2004 05:46 PM

14 comments, last by vaneger 20 years ago

yspotua

122

April 03, 2004 06:11 PM

quote:Original post by vaneger
actually i goofed i dont need numbers that multiply to get a producti need numbers that add to get a number how can i find that? such as 2 + 5 = 7 or 3 + 4 = 7 etc

Assuming that you are dealing with integers: start with the number, subtract one. The result and 1 is one set.

Then subtract 2, the result and 2 is another set. And so on and so forth.

BTW.

I dont't think the OP asked for prime factors anyways.

[edited by - yspotua on April 3, 2004 8:52:19 PM]

zealouselixir

256

April 03, 2004 06:25 PM

Aprosenf: Hm?
>> 216
2 2 2 3 3 3

Looks like it works to me. What number would break it?

[twitter]warrenm[/twitter]

python_regious

929

April 03, 2004 07:05 PM

quote:Original post by vaneger
actually i goofed i dont need numbers that multiply to get a producti need numbers that add to get a number how can i find that? such as 2 + 5 = 7 or 3 + 4 = 7 etc

There are an infinite number solutions to that problem, you need to add some constraints.

You have to remember that you''re unique, just like everybody else.

If at first you don't succeed, redefine success.

vaneger

100

Author

April 03, 2004 08:27 PM

no there are 8 solutions to that problem i believe
0+7 1+6 2+5 3+4 7+0 6+1 5+2 4+3 or only 4 if order is limited.

intrest86

742

April 03, 2004 09:45 PM

No, he was right to say there are infinite answers. For example, 100 + (-93) = 7. Now, if you add some constraints like he said...

Turring Machines are better than C++ any day ^_~

vaneger

100

Author

April 03, 2004 10:01 PM

whole positive numbers less than or equal to the given number
the number of additive pairs is equal to (n+1)/2 with the above example of 7 the number of pairs is 4 (07,16,25,34) not counting reverses. i currently have this code

struct pr{	int first;	int second;};vector<pr> pairs;void set_pairs(int n){	int q = (n+1) / 2;	pairs.resize(q)	for(int i = 0; i < q;i++)	{		pairs[i].first = n -  i;		pairs[i].second = n - (n-i); 	}}

but im not sure if set_pairs can be made faster or not.

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