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.
1 #include <stdio.h> 2 #include <string.h> 3 #include <ctype.h> 4 5 void solve() 6 { 7 int sum,i; 8 sum=0; 9 for(i=3; i<1000; i++) 10 { 11 if(i%3==0 || i%5==0) 12 { 13 sum+=i; 14 } 15 } 16 printf("%d\n",sum); 17 18 } 19 20 int main() 21 { 22 solve(); 23 return 0; 24 }
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Answer:
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233168 |