// ////////////////////////////////////////////////////////
// # Title
// Multiples of 3 and 5
//
// # URL
// https://projecteuler.net/problem=1
// http://euler.stephan-brumme.com/1/
//
// # Problem
// 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.
//
// # Solved by
// Stephan Brumme
// February 2017
//
// # Algorithm
// We are supposed to find of all multiples of 3 or 5 __below__ the input number,
// therefore we decrement it by one.
//
// In general, the sum of all numbers between 1 and `x` is `sum_{1..x}i=x * (x+1)/2`
// (see https://en.wikipedia.org/wiki/Triangular_number )
//
// There are `floor{x/3}` numbers between 1 and x which are divisible by 3 (assuming `floor{x/3}` is an integer division).
// e.g. the range `1..10` contains `floor{10/3}=3` such numbers (it's 3, 6 and 9). Their sum is `3+6+9=18`.
// This can be written as `3/3 * (3+6+9)` which is the same as `3 * (3/3+6/3+9/3)=3 * (1+2+3)`.
// Those brackets represent ` sum_{1..3}i = sum_{1..10/3}i` (or short: ` sum{10/3}`)
// and thus our overall formula for the sum of all multiples of 3 becomes `3 * sum{x/3}`.
//
// The same formula can be used for 5:
// The sum of all numbers divisible by 5 is `5 * sum{x/5}`
//
// However, there are numbers divisible by 3 __and__ 5, which means they are part of __both__ sums.
// We must not count them twice, that's why we (in addition to the aforementioned sums)
// compute the sum of all numbers divisible by 3*5=15 to correct for this error.
//
// In the end we print ''sumThree + sumFive - sumFifteen''
//
// # Alternative
// Looping through all numbers from 1 and 1000 and checking each of those numbers
// whether they are divisible by 3 or 5 easily solves the problem, too, and produces the result pretty much instantly.
// Even more, the code will be probably a bit shorter.
//
// However, Hackerrank's input numbers are too large for that simple approach (up to `10^9` with `10^5` test cases)
// and will lead to timeouts.

#include <iostream>

// triangular number: `sum{x}=1+2+..+x = x*(x+1)/2`
unsigned long long sum(unsigned long long x)
{
  return x * (x + 1) / 2;
}

int main()
{
  unsigned int tests;
  std::cin >> tests;
  while (tests--)
  {
    unsigned long long last;
    std::cin >> last;

    // not including that number
    last--;

    // find sum of all numbers divisible by 3 or 5
    auto sumThree   =  3 * sum(last /  3);
    auto sumFive    =  5 * sum(last /  5);

    // however, those numbers divisible by 3 AND 5 will be counted twice
    auto sumFifteen = 15 * sum(last / 15);

    std::cout << (sumThree + sumFive - sumFifteen) << std::endl;
  }

  return 0;
}