For anyone who is using Python3. If you've already converted from the iterive to the O(1) solution but you're still getting the "wrong answer" with Test Cases 2 and 3 here is why:

When you divide integers in python 3, the answer is converted automatically to a float. Becuase Test Cases 2 and 3 involve very large numbers, when you convert back to integer, there will be a huge rounding error and your answers will be wrong.

This is the issue I had to resolve. Luckily for me, my only division was a divide by two so I converted it to a bitwise right-shift.

O(1) Solution : say N= 100 so, max multiple of 3 here below N is 99 i.e. 3*33 so, sum of all multiples of 3 has a pattern 3(1+2+3+....+33) use this

Here is the code in C++

long N,num,three,five,fifteen=0;
cin>>N;

for(int i=0;i<N;i++)
{
    cin>>num;
    //int sum=0;
    three=(num-1)/3;
    five=(num-1)/5;
    fifteen=(num-1)/15;

    cout << 3*(three*(three+1)/2)+5*(five*(five+1)/2)-15*(fifteen*(fifteen+1)/2)<<endl;

}
return 0;

The first challenge and I'm already wishing I'd paid more attention in math class.

Here's how I approached the problem.( I used Java ) First I scanned akk the numbers and checked using modulus. This gave me a time out error. Then I used the sum of arithmatic progression formula, which got past the tine out error , but gave me a wrong answer error. This was because, the sums generated in test case 3 and 4 results were out of range of the long number's range. So, I used the BigInteger for sums and long for inputs. And finally the solution worked.

If you get an timeout in case 2 and 3, stop iterating and checking with modulo. You need to calculate it, otherwise it's too slow.