Re: Math Question....

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On 22 April 2010 15:26, Richard Quadling <rquadling@xxxxxxxxxxxxxx> wrote:
> On 22 April 2010 15:13, Ashley Sheridan <ash@xxxxxxxxxxxxxxxxxxxx> wrote:
>> On Thu, 2010-04-22 at 10:17 -0400, Dan Joseph wrote:
>>
>>> On Thu, Apr 22, 2010 at 10:12 AM, Stephen <stephen-d@xxxxxxxxxx> wrote:
>>>
>>> > 1,252,398 DIV 30 = 41,746 groups of 30.
>>> >
>>> > 1,252,398 MOD 30 = 18 items in last group
>>> >
>>> Well, the only problem with going that route, is the one group is not
>>> equally sized to the others.  18 is ok for a group in this instance, but if
>>> it was a remainder of only 1 or 2, there would be an issue.  Which is where
>>> I come to looking for a the right method to break it equally.
>>>
>>
>>
>> How do you mean break it equally? If the number doesn't fit, then you've
>> got a remainder, and no math is going to change that. How do you want
>> that remainder distributed?
>>
>> Thanks,
>> Ash
>> http://www.ashleysheridan.co.uk
>>
>>
>>
>
> It sounds like you are looking for factors.
>
> http://www.algebra.com/algebra/homework/divisibility/factor-any-number-1.solver
>
> Solution by Find factors of any number
>
> 1252398 is NOT a prime number: 1252398 = 2 * 3 * 7 * 29819
> Work Shown
>
> 1252398 is divisible by 2: 1252398 = 626199 * 2.
> 626199 is divisible by 3: 626199 = 208733 * 3.
> 208733 is divisible by 7: 208733 = 29819 * 7.
> 29819 is not divisible by anything.
>
> So 29819 by 42 (7*3*2)
>
> would be a route.
>
>
> Take note of http://www.algebra.com/algebra/homework/divisibility/Prime_factorization_algorithm.wikipedia,
> which has the comment ...
>
> "Many cryptographic protocols are based on the difficultly of
> factoring large composite integers or a related problem, the RSA
> problem. An algorithm which efficiently factors an arbitrary integer
> would render RSA-based public-key cryptography insecure.".
>
>
>
>
> --
> -----
> Richard Quadling
> "Standing on the shoulders of some very clever giants!"
> EE : http://www.experts-exchange.com/M_248814.html
> EE4Free : http://www.experts-exchange.com/becomeAnExpert.jsp
> Zend Certified Engineer : http://zend.com/zce.php?c=ZEND002498&r=213474731
> ZOPA : http://uk.zopa.com/member/RQuadling
>

Aha. Missed the "30" bit.

So, having found the factors, you would need to process them to find
the largest combination under 30.

2*3
2*3*7
2*7
3*7

are the possibilities (ignoring any number over 30).

Of which 3*7 is the largest.

So, 1,252,398 divided by 21 = 59,638


Is that the sort of thing you are looking for?

-- 
-----
Richard Quadling
"Standing on the shoulders of some very clever giants!"
EE : http://www.experts-exchange.com/M_248814.html
EE4Free : http://www.experts-exchange.com/becomeAnExpert.jsp
Zend Certified Engineer : http://zend.com/zce.php?c=ZEND002498&r=213474731
ZOPA : http://uk.zopa.com/member/RQuadling

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