From: "Joe Wölfel" <joe@...>
Date: 2008-10-18T23:17:48+09:00
Subject: Re: Counting

Don't need no stinkin' select either;)  It's much faster if you don't  
iterate at all.

n=1000
d = 10
(n+d)*(n/(d*2))
=> 50500

The formula is all you need because
1000 + 10 = 1010
990 + 20 = 1010
etc.  50 times.

The following select gives the same answer but is much slower.
(1..n).select{|a| 0 == a % d }.inject {|sum, x| sum+x}
=> 50500

One caveat.  The formula I gave assumes that 'n' is divisible by  
'd'.  It shouldn't be hard to change the formula so that isn't  
necessary though.

Cheers,
Joe



On 18 oct. 08, at 08:39, William James wrote:

> Tom Clarke wrote:
>
>> How would i go about making Ruby count to say 1000 usin only  
>> multiples
>> of say 2 and 6. Then i am looking to add all of the outputted  
>> numbers.
>> Can anyone help
>
> "We don't need no stinkin' loops!"
>
> (1..1000).select{|n| 0 == n % 10 }
>     ==>
> [10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140,
> 150, 160, 170, 180, 190, 200, 210, 220, 230, 240, 250, 260, 270,
> 280, 290, 300, 310, 320, 330, 340, 350, 360, 370, 380, 390, 400,
> 410, 420, 430, 440, 450, 460, 470, 480, 490, 500, 510, 520, 530,
> 540, 550, 560, 570, 580, 590, 600, 610, 620, 630, 640, 650, 660,
> 670, 680, 690, 700, 710, 720, 730, 740, 750, 760, 770, 780, 790,
> 800, 810, 820, 830, 840, 850, 860, 870, 880, 890, 900, 910, 920,
> 930, 940, 950, 960, 970, 980, 990, 1000]
>
> -- 
>
>