On Sep 4, nyec said: >The program took a number sequence. For example <10-100>, reversed the digits >and calculated the difference of the original number. So the calculations >look like this: <10-01=9; 11-11=0; 12-21=9; 13-31=18; etc.>.
A number subtracted from its reverse yields a multiple of 9. Notice that 52 - 25 = 27, which is 3*9. Moreover, 3 = 5 - 2 (the digits of the number). Notice that 461 - 164 = 297, which is 3*99. Moreover, 3 = 4 - 1. Notice that 4321 - 1234 = 3087, which is (whew!) 3*999 + 1 * 90. Moreover, 3 = 4 - 1, and 1 = 3 - 2. Do you see the pattern here? It's rather easy to prove mathematically: AB - BA = 10A + B - 10B - A = 9A - 9B = 9*(A-B) which must be divisible by 9, and so on. -- Jeff "japhy" Pinyan [EMAIL PROTECTED] http://www.pobox.com/~japhy/ RPI Acacia brother #734 http://www.perlmonks.org/ http://www.cpan.org/ ** Look for "Regular Expressions in Perl" published by Manning, in 2002 ** <stu> what does y/// stand for? <tenderpuss> why, yansliterate of course. [ I'm looking for programming work. If you like my work, let me know. ] -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
