Original Post
Challenge
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.
This is project euler challenge 1. I'd like to see how many languages we can write a solution in.
I'll write a few and post them here soon. This is for fun, so make your code as short or as efficient as you can.
PHP Code:
function all_multiples($num) {
$arr = array();
for ($i = $num; $i < 1000; $i += $num)
$arr[] = $i;
return $arr;
}
$result = array_map('all_multiples', array(3, 5));
$end_result = array_unique(array_merge($result[0], $result[1]));
sort($end_result);
echo(array_sum($end_result));
sum [x | x <- [1 .. 999], x `mod` 3 == 0 || x `mod` 5 == 0]Ruby code:
total = 0
(1 .. 999).each { |i|
if (i % 3 == 0 or i % 5 == 0) then
total += i
end
}
Last edited by Ryan_old2; May 23, 2011 at 03:07 PM.
C# code:
C++ code:
C code:
Java code:
VB code:
F# code:
didn't test any of them lol \o/
You know, it would probably be more efficient to do something like;
(int)999/3 + (int)999/5 - (int)999/15
rather than looping (didn't check my maths, but it seems legit to me)
int total = 0;
for(int i=1;i<1000;i++) total+=(i%3==0||i%5==0)?1:0;
return total;C++ code:
int total = 0;
for(int i=1;i<1000;i++) if(i%3==0||i%5==0) total++;
return total;C code:
int total = 0;
for(int i=1;i<1000;i++) if(i%3==0||i%5==0) total++;
return total;Java code:
int total = 0;
for(int i=1;i<1000;i++) if(i%3==0||i%5==0) total++;
return total;VB code:
Dim total = 0
For i = 1 To 999 Step 1
If ((i mod 3)=0 Or (i mod 5)=0) Then
total = total+1
End IF
Next i
Return totalF# code:
let mutable total = 0
let pool={1..1000}
let filter =
pool
|> Seq.filter (fun x -> i%3 = 0 || i%5 = 0)
|> Seq.map (fun x -> total <- total + 1)
return totaldidn't test any of them lol \o/
You know, it would probably be more efficient to do something like;
(int)999/3 + (int)999/5 - (int)999/15
rather than looping (didn't check my maths, but it seems legit to me)
Ruby code:
My version of the ruby solution.
I'll do lua later.
number=0
for i in 1..999
if i%3==0 or i%5==0
number=number+i
end
end
puts number.to_s
My version of the ruby solution.
I'll do lua later.
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javascript code:
Also, the following function ported to any language and called with parameters 3 and 5:
var total = 0;
for(var i = 0; i < 1000; i++) {
if(i % 3 == 0 || i % 5 == 0) {
total += i;
}
}
alert(total);Also, the following function ported to any language and called with parameters 3 and 5:
function f(n1, n2) {
return n1*((1000/n1) * ((1000/n1)+1) / 2) + n2*((1000/n2) * ((1000/n2)+1) / 2) - (n1*n2)*((1000/(n1*n2) * ((1000/(n1*n2))+1) / 2);
}
Last edited by Ryan_old2; May 23, 2011 at 03:59 PM.
Also use 4 spaces = 1 tab, that is the standard.