Re: Math Question....

On 22 April 2010 15:13, Ashley Sheridan <ash@ashleysheridan.co.uk> 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@rogers.com> 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