My 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 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 Approaches

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.