Re: Math Question....

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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.".




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