Oh yes, BTW, if you allow zero prefixes on your numbers, there are solutions with multiplier 99, 999, etc. For example, the smallest solution for 99 is this monster: 99*(10^486-1)/9899 = 01000101020305081321345590463683200323264976260228305889483786241034 44792403273057884634811597131023335690473785230831397110819274674209 51611273866047075462167895747045156076371350641478937266390544499444 38832205273259925244974239822204263056874431760783917567431053641781 99818163450853621577937165370239418123042731589049398929184766137993 73674108495807657339125164157995757147186584503485200525305586422870 997070411152641680977876553187190625315688453379129204970199 but we need that initial 0 to make it work. Also note initial elements of the Fibonacci sequence hiding near the beginning of the number. This is a feature of this sort of number. ----- Original Message ----- From: "msubbara" <msubbara@ualberta.ca> To: <ham>; "Math Fun" <math-fun@mailman.xmission.com>; "Richard Guy" <rkg@cpsc.ucalgary.ca>; "seqfan" <seqfan@ext.jussieu.fr> Sent: Thursday, April 28, 2005 7:35 PM Subject: [math-fun] RE: Unsolved? problem
What would be the solution if we do similar operation with the last two digits and ask '99 times the original number", or replace the last three digits similarly to get 999 times the original number etc. Subbarao