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

# Counting

**tbcox23**#1

**tbcox23**#2

Here’s one way:

multiples = []

(1…1000).each do |number|

if (number % 2) == 0 && (number % 6) == 0

multiples << number

end

end

puts “multiples of 2 and 6 between 1 and 1000: #{multiples.join(’, ')}”

print "sum of multiples: "

puts multiples.inject(0) { |sum, multiple| sum + multiple }

Regards,

Craig

**tbcox23**#3

Can you elaborate a bit more?

Are you looking for all multiples of 2 and 6 under 1000?

Mathematically,

that’s just all multiples of 6 under 1000, so you could do something

like

n=6

i = 1

array = []

while n*i < 1000 do
array.push(n*i)

i += 1

end

p array

I’m not sure what you’re asking, but maybe if you can give an example of

what you’re looking for I can be more useful.

On Fri, Oct 17, 2008 at 4:44 PM, Tom Clarke

**tbcox23**#4

On Fri, Oct 17, 2008 at 4:07 PM, Craig D.

removed_email_address@domain.invalidwrote:

puts multiples.inject(0) { |sum, multiple| sum + multiple }

Regards,

Craig

You should be able to ignore the % 2 bit as if it’s divisible by 6 then

it’s

divisible by 2

–

“Hey brother Christian with your high and mighty errand, Your actions

speak

so loud, I can’t hear a word you’re saying.”

-Greg Graffin (Bad Religion)

**tbcox23**#6

On Oct 17, 2008, at 6:31 PM, Tom Clarke wrote:

Sorry i meant to say 2 and 5

You should just have the professor post the assignment directly and it

would avoid such confusion.

-Rob

Rob B. http://agileconsultingllc.com

removed_email_address@domain.invalid

**tbcox23**#7

Using the fact that a multiple of 6 must be a multiple of 2 - apart from

2

and 4:

(6…1000).inject(0) {|sum, i| i % 6 == 0 ? sum + i : sum} + 6

On Fri, Oct 17, 2008 at 4:44 PM, Tom Clarke

**tbcox23**#8

Rob B. wrote:

On Oct 17, 2008, at 6:31 PM, Tom Clarke wrote:

Sorry i meant to say 2 and 5

You should just have the professor post the assignment directly and it

would avoid such confusion.-Rob

Rob B. http://agileconsultingllc.com

removed_email_address@domain.invalid

what youmean the proffesor

**tbcox23**#9

On Fri, Oct 17, 2008 at 4:21 PM, Tom Clarke

removed_email_address@domain.invalid wrote:

Rob B. http://agileconsultingllc.com

removed_email_address@domain.invalidwhat youmean the proffesor

Professor: A teacher or instructor usually in a class room environment

who hands out homework assignments asking for arbitrary and generally

worthless tasks to be completed like the one you have posted.

Ie: If you want us to do your homework for you, just paste in the

question next time.

**tbcox23**#10

I like that approach, William. Thanks for the deodorant.

Craig

P.S. The OP wanted n % 6, not n % 10.

**tbcox23**#11

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]

**tbcox23**#12

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

**tbcox23**#13

# The OP needs multiples of 2 and 6. Since all multiples of 6 are also

multiples of 2, just checking for multiples of 6 suffices. However, if d

6, your two approaches don’t produce the same result. For d = 6, they

produce 83498 and 83166, respectively. 83166 is the correct answer, I

believe.

Regards,

Craig

**tbcox23**#14

On Sat, Oct 18, 2008 at 3:30 PM, Todd B. removed_email_address@domain.invalid

wrote:

I think he meant (multiples of 2) AND (multiples of 6), or in this

case (multiples of 2) AND (multiples of 5)

With ‘he’, I meant the OP.

Todd

**tbcox23**#15

On Fri, Oct 17, 2008 at 5:44 PM, Tom Clarke

removed_email_address@domain.invalid 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

FizzBuzz FAIL!

**tbcox23**#16

On Sat, Oct 18, 2008 at 10:02 AM, Craig D.

removed_email_address@domain.invalid wrote:

The OP needs multiples of 2 and 6. Since all multiples of 6 are also

multiples of 2, just checking for multiples of 6 suffices. However, if d =

6, your two approaches don’t produce the same result. For d = 6, they

produce 83498 and 83166, respectively. 83166 is the correct answer, I

believe.

Well, I don’t think he meant all multiples of (2 AND 6), or in this

case (2 AND 5). That will always be simply always be multiples of the

LCM, which, for (2 AND 6) is 6, for (12 AND 18) is 36, for (2 AND 5),

10, etc.

lcm = 2.lcm 5

puts (1…1000/lcm).map.inject {|s, i| s + i * lcm}

I think he meant (multiples of 2) AND (multiples of 6), or in this

case (multiples of 2) AND (multiples of 5), which would be multiples

of either. In that case check whether the number is cleanly divisible

by number a and number b using %.

Todd

**tbcox23**#17

Right, as per my caveat n must be divisible by d*2. You can’t

divide 1000 by twelve. But 996 is divisible by 12. So you can use

that for d instead for that particular case. If you played around

with remainders for a bit you could adjust the formula I gave to work

for all cases.

The point is you go from linear time for all the iterative methods to

constant time if you just work out a formula. For very large numbers

the iterative methods will take more time than you are willing to

spend and perhaps never finish at all.

**tbcox23**#18

Tom Clarke wrote:

of say 2 and 6. Then i am looking to add all of the outputted numbers.

Can anyone help

If you want numbers that are divisible by 2 *or* 6, then that is all

even numbers. If you want numbers that are divisible by 2 *and* 6, then

that you really need only to check multiples of 6. Why can’t you use

something like this:

6.step(1000, 6)

You need only start with 6 to find multiples. Every sixth number will

be a multiple.