# Problem 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 1000.

# 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)

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

       #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;
}


This solution contains 7 empty lines, 4 comments and 1 preprocessor command.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 100" | ./1

Output:

Note: the original problem's input 1000 cannot be entered
because just copying results is a soft skill reserved for idiots.

(this interactive test is still under development, computations will be aborted after one second)

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=c++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

# Changelog

February 22, 2017 submitted solution

# Hackerrank

My code solved 5 out of 5 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 5% (out of 100%).

Hackerrank describes this problem as easy.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is never an option.

projecteuler.net/thread=1 - the best forum on the subject (note: you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-problem-1/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p001.java (written by Nayuki)
Mathematica: github.com/nayuki/Project-Euler-solutions/blob/master/mathematica/p001.mathematica (written by Nayuki)
C: github.com/eagletmt/project-euler-c/blob/master/1-9/problem1.c (written by eagletmt)
Go: github.com/frrad/project-euler/blob/master/golang/Problem001.go (written by Frederick Robinson)
Javascript: github.com/dsernst/ProjectEuler/blob/master/1 Multiples of 3 and 5.js (written by David Ernst)
Scala: github.com/samskivert/euler-scala/blob/master/Euler001.scala (written by Michael Bayne)

# Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.
yellow problems score less than 100% at Hackerrank (but still solve the original problem).
gray problems are already solved but I haven't published my solution yet.
blue problems are already solved and there wasn't a Hackerrank version of it (at the time I solved it) or I didn't care about it because it differed too much.

Please click on a problem's number to open my solution to that problem:

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The 133 solved problems had an average difficulty of 16.9% at Project Euler and I scored 11,174 points (out of 12300) at Hackerrank's Project Euler+.
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