Project Euler #1: Multiples of 3 and 5

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 N .

Input Format
First line contains T that denotes the number of test cases. This is followed by T lines, each containing an integer, N .

Output Format
For each test case, print an integer that denotes the sum of all the multiples of 3 or 5 below N .

Constraints
1≤T≤105
1≤N≤109

Sample Input

2
10
100

Sample Output

23
2318 
  
  
  
  
                                    
   
   
   
   
 
    
Language: C
 
   
                                  
   
   
   
   
 
   
                            
  
  
  
  
            
  
  
  
  

   
  
  
  
  
  
  
  
  

   
   
   
   
 
    
 
     
 
      
 
       
 
      
 
      
 
       
 
        
 
         

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#include <stdio.h>
 
       
 
       
 
        
 
         

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#include <stdlib.h>
 
       
 
       
 
        
 
         

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​
 
       
 
       
 
        
 
         

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int main()
 
       
 
       
 
        
 
         

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{
 
       
 
       
 
        
 
         

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    unsigned long long int N,i,j,x,y,z;
 
       
 
       
 
        
 
         

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    int T;
 
       
 
       
 
        
 
         

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    scanf("%d",&T);
 
       
 
       
 
        
 
         

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    unsigned long long int n[T];
 
       
 
       
 
        
 
         

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    for(i=1;i<=T;i++)
 
       
 
       
 
        
 
         

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    {
 
       
 
       
 
        
 
         

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        scanf("%llu",&N);
 
       
 
       
 
        
 
         

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        x=(N-1)/3;
 
       
 
       
 
        
 
         

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        y=(N-1)/5;
 
       
 
       
 
        
 
         

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        z=y/3;
 
       
 
       
 
        
 
         

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        n[i-1]=(3+3*x)*x/2;
 
       
 
       
 
        
 
         

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        n[i-1]+=(5+5*y)*y/2;
 
       
 
       
 
        
 
         

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        n[i-1]-=(15+15*z)*z/2;
 
       
 
       
 
        
 
         

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    }
 
       
 
       
 
        
 
         

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    for(i=1;i<=T;i++)
 
       
 
       
 
        
 
         

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    {
 
       
 
       
 
        
 
         

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        printf("%llu\n",n[i-1]);
 
       
 
       
 
        
 
         

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    }
 
       
 
       
 
        
 
         

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    return 0;
 
       
 
       
 
        
 
         

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}