On 22 April 2010 15:26, Richard Quadling <rquadl...@googlemail.com> wrote: > On 22 April 2010 15:13, Ashley Sheridan <a...@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 <stephe...@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. Â 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. Â 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 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
