Finding divisors

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?