[Gloin]

Are you just referring to the 8 syllables or were you looking for something else?

[BigAnt]

It was something else.....try counting.

[BigAnt]

Well here is the solution:

Solution:
3,1 4 1 5 9
2 6 5 3 5 8
9 7 9
3 2 3 8 4 6

That's Pi!

Let's try this one:

As I was going to Saint Ives,
I crossed the path of seven wives.
Every wife had seven sacks,
Every sack had seven cats,
Every cat had seven kittens,
Kittens, cats, sacks, wives,
How many were going to Saint Ives?

[BigAnt]

Anyone?

[krzysz00]

KYA, on 22 Dec, 2008 - 05:05 PM, said:

OK. Next math problem:

File's attached. Paint was being retarded. It's a series.

How DARE you use proprietary formats for reasons other than the fact that the recipient requests it/demands it/thinks that the format's software is better and refuses to install the free alternative. (Like my mom who refuses to try OpenOffice even though it's perfectly fine). Many of us know there are freely avalable alternatives to M$ Word. (If I am ranting, I forgot I wasn't on a Linux forum).

[BigAnt]

Quote

How DARE you use proprietary formats for reasons other than the fact that the recipient requests it/demands

Then try my problem as no format is needed.

[BigAnt]

Apparently this thread has died, so while my previous questions lies in purgatory waiting for a answer I shall give a mini question to revive the thread:

What is the smallest integer greater than 0 that can be written entirely with zeros and ones and is evenly divisible by 225?

[Gloin]

Unless you were thinking of the binary number 11100001
I would say 11,111,111,100

[BigAnt]

Quote

Unless you were thinking of the binary number 11100001
I would say 11,111,111,100

And we've got a winner!

(.) (.) (.) (.) (.) <----Cookies for you.

This post has been edited by BigAnt: 09 April 2009 - 05:40 PM

[Gloin]

Wohoo!! You know how to please me..

For those who wonder how I came up with the answer, a short explanation..

First I calculated the prime factors of 225 = 9 * 5 * 5.
So the number to find had to be divisible by both 9 and 25.
The only way a number that consists of only 0's and 1's could be divisible by 9 is if it contains an exact multiple of nine 1's (and arbitrary many 0's) because if a number is divisible by nine, then the sum of the digits has to be a multiple of 9.
Ex: 45 is divisible by 9 and 4+5 = 9

The smallest number containing nine 1's is 111,111,111 but that is not divisible by 25. In fact, any number containing only 0's and 1's that is divisible by 25 must also be divisible by 100. Therefor I multiplied the nine 1's by 100 and thus came up with the answer. Magic!!


I'm gonne try to come up with a new problem but if anyone else has one, feel free to post it.

[runfaster]

@BigAnt:

In reply to your first question, ONE is going to St. Ives. Die Hard with a Vengence FTW.

[Gloin]

Linear Feedback Shift Register

Find the (shortest) initializing bitstring and the recursive function that produces the bitstring,

1001110 1001110 1001110 ... (continues with the same pattern)

(modulo-2 arithmetics apply without carry)

[BigAnt]

Quote

@BigAnt:

In reply to your first question, ONE is going to St. Ives. Die Hard with a Vengence FTW.

Yep

[firebolt]

An integer minus another, divided by, the second integer minus the first gives what??

Its pretty easy...

This post has been edited by firebolt: 10 April 2009 - 12:11 AM

[Gloin]

Firebolt>> -1

In my question I'm looking for a specific bitstring of a specific length and a recursive function that reproduces the bitstring I posted.

ex:

Bitstring: 1101 where [x(0) = 1, x(1) = 1, x(2) = 0, x(3) = 1]
Recursive function: x(2) = x(1) + x(0)
Would produce the bitstring: 110 110 110 110 ...

This post has been edited by Gloin: 10 April 2009 - 06:57 AM