Finding divisors

Ok the divisors of 10 are 1 , 2 , 5 , 10 . Ok I'm making a program, but I can't think of a way to find the divisor of a number. Example if someone types 6 how do I find it's divisor? I just need help on te math portion.

Note: I'm not asking people to do this code for me I just need some help on finding the divisor of a number.

Do you know how to build a factor tree?

You can do this recursively. If it is even, divide in half, and factor again. If the number is odd, check for prime, otherwise factor the composite.

1 and the integer are given factors of any integer.

Quote:

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Originally Posted by citizen

Do you know how to build a factor tree?

You can do this recursively. If it is even, divide in half, and factor again. If the number is odd, check for prime, otherwise factor the composite.

1 and the integer are given factors of any integer.

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so num /= 2?

You can do a very basic type of proceedure for brute forcing it.

Example:

Code:

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int nfactors(int number, std::vector<int> &factors)
{
  int n = 0, i = 1, limit = number;

  factors.clear();

  for(;i < limit; ++i)
    if(!(number&#37;i))
    {
      limit = number/i;
      factors.push_back(limit);
      ++n;

      if(limit != i)
      {
          factors.push_back(i);
          ++n;
      }
    }

  // One could sort the results at this point, but bleh... you didn't specify that.

  return n;
}

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Quote:

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Originally Posted by master5001

You can do a very basic type of proceedure for brute forcing it.

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n seems rather redundant since there is factors.size().

True.... I have no explanation for that other than I have been getting less and less sleep each passing day. Though, that is hardly a valid excuse for a programmer.

Are you trying to reduce fractions, xbusterx? You could always use the Euclidean Algorithm. That would be a much more apt method for finding a greatest common factor.

Quote:

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Originally Posted by citizen

Do you know how to build a factor tree?

You can do this recursively. If it is even, divide in half, and factor again. If the number is odd, check for prime, otherwise factor the composite.

1 and the integer are given factors of any integer.

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I know that but even numbers don't just divide by 2 , they divide by 3 , 5 , 5 etc as well. I know I have to use num % n1 = 0 . But ya I'm lost on the much part plz help.

Quote:

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Originally Posted by xbusterx

I know that but even numbers don't just divide by 2 , they divide by 3 , 5 , 5 etc as well. I know I have to use num % n1 = 0 . But ya I'm lost on the much part plz help.

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So check all the numbers. num%2, num%3, num%4, ....

Quote:

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Originally Posted by tabstop

So check all the numbers. num%2, num%3, num%4, ....

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ok what if I say this:

Code:

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int n;
cout << enter number;
cin >> n;

for (int i = 1 ; i < n ; i++ )

if ( n % i = 0 )

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how do I extract the divisor?

Quote:

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Originally Posted by xbusterx

ok what if I say this:

Code:

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int n;
cout << enter number;
cin >> n;

for (int i = 1 ; i < n ; i++ )

if ( n % i = 0 )

---

how do I extract the divisor?

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Divide?

Quote:

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Originally Posted by tabstop

Divide?

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No that's not what I mean. IF ( n % i = 0 ) that means i is a divisor, but how do I extract the divisors so I can add them? sorry I'm new.

Quote:

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Originally Posted by xbusterx

No that's not what I mean. IF ( n % i = 0 ) that means i is a divisor, but how do I extract the divisors so I can add them? sorry I'm new.

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You don't have to extract any divisors -- you just said that i was a divisor. So that's the number you have to add.

Quote:

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Originally Posted by tabstop

You don't have to extract any divisors -- you just said that i was a divisor. So that's the number you have to add.

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exactly you're right , BUT how do I add them? that's why I need to exact the divisors to add them.

Quote:

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Originally Posted by xbusterx

exactly you're right , BUT how do I add them? that's why I need to exact the divisors to add them.

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I usually add numbers with "+".

I mean, you know what a running total is, right?

Quote:

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Originally Posted by tabstop

I usually add numbers with "+".

I mean, you know what a running total is, right?

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um like sum += num ?

Quote:

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Originally Posted by xbusterx

um like sum += num ?

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Right.

Quote:

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Originally Posted by tabstop

Right.

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Ya don't I don't know how to do that here since i is changing.

Quote:

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Originally Posted by xbusterx

Ya don't I don't know how to do that here since i is changing.

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Since that's exactly what you want and need to happen, I fail to see how that's a problem.

(Or is your task not to add, but to count? In that case you would always add 1, since that's how people count.)

Quote:

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Originally Posted by tabstop

Since that's exactly what you want and need to happen, I fail to see how that's a problem.

(Or is your task not to add, but to count? In that case you would always add 1, since that's how people count.)

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Ok so can I do this?

Code:

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int sum =0;
int n;

cout << enter number;
cin >> n;

for (int i = 1 ; i < n ; i++ )

if ( n &#37; i = 0 )
    sum += i

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You have to use == to compare numbers, but yes, that's what we've been waiting for.

Quote:

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Originally Posted by tabstop

You have to use == to compare numbers, but yes, that's what we've been waiting for.

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Code:

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int sum =0;
int n;

cout << enter number;
cin >> n;

for (int i = 1 ; i < n ; i++ )

if ( n % i == 0 )
    sum += i

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better lol?