## Project Euler Problem 1

### February 10, 2015

Project Euler is a collection of math problems intended for computer solution, and is one of the inspirations for Programming Praxis. The first problem on Project Euler, which also regularly appears on lists of phone interview questions, asks you to:

Find the sum of all the multiples of 3 or 5 below 1000.

Your task is to write a program that solves Problem 1 for arbitrary *n*; since the problem is simple, you must provide at least three fundamentally different solutions. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

Advertisements

Pages: 1 2

Three ways…

(1) Brute force overall entries and summing them!

(2) Brute force summing muliples of 5, multiples of 3 and removing duplicate multiples of 15…

(3) Same as 2 but using formulae to compute sum – no need to brute force…

sub bf {

my $n = shift;

my $t = 0;

$t+= $_ foreach grep { !($_ % 3) || !($_ % 5) } 1..($n-1);

return $t;

}

sub x {

my $n = shift;

return _x($n,3)+_x($n,5)-_x($n,15);

}

sub _x {

my($n,$f)=@_;

my $s = 0;

my $t = 0;

while($t<$n) {

$s+=$t;

$t+=$f;

}

return $s;

}

sub yy {

my $n = shift;

return _y($n,3)+_y($n,5)-_y($n,15);

}

sub _y {

my($n,$f)=@_;

$n = int (($n-1)/$f);

return $f * ($n+1)*$n/2;

}

printf "%7d %20d %20d %20d\n", $_, bf($_), x($_), yy($_) foreach @ARGV;

My solution in Python.

Easy to extend, just add your own function as long as it starts with “solve_” and returns the result as an int.

Above code outputs for n = 1000:

Running method: solve_brute_force

233168

Run time (s): 0.0

Running method: solve_plain_math

233168

Run time (s): 0.0

Running method: solve_wheel_multiples

233168

Run time (s): 0.0

——————————————————–

For n = 10000000:

Running method: solve_brute_force

23333331666668

Run time (s): 1.47399997711

Running method: solve_plain_math

23333331666668

Run time (s): 0.0

Running method: solve_wheel_multiples

23333331666668

Run time (s): 0.986999988556

Similar solution using Gauss’s summation formula in SML:

Only two ways but they are more different. I suppose the sampling method

is O(sample size) space, while the gambling method is O(1) space, but the

sample size can be considered a constant. These count 0 as a multiple.