Project Euler - Problem 001

August 1st, 2007 by Daniel Høyer Iversen

If we list all the natural numbers below
10 that are multiples of 3 or 5, we get 3, 5,
6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

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#include <iostream> ;
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using namespace std;
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int main ()
 
{
 
int res = 0;
 
int max = 1000;
 
int multipleOfThree;
 
int multipleOfFive;
 
for (int i=1; 3*i < max; i++)
 
{
 
multipleOfThree=3*i;
 
multipleOfFive=5*i;
 
if (multipleOfThree >= 1000)
 
multipleOfThree=0;
 
if (multipleOfFive >= 1000)
 
multipleOfFive=0;
 
if(multipleOfThree%5==0)
 
multipleOfThree=0;
 
res += multipleOfThree+multipleOfFive;
 
}
 
cout << "resultatet er ";
 
cout << res;
 
return 0;
 
}
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