Even Fibonacci numbers - problem 2 >> |

# Problem 1: Multiples of 3 and 5

(see projecteuler.net/problem=1)

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)

and will lead to timeouts.

# 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:

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

March 23, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler001

My code solves **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.

# Links

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)

Haskell: github.com/nayuki/Project-Euler-solutions/blob/master/haskell/p001.hs (written by Nayuki)

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 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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My username at Project Euler is

**stephanbrumme**while it's stbrumme at Hackerrank.

Even Fibonacci numbers - problem 2 >> |