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
If you take n digits from the right of a number and prepend them back, the result has the same number of digits as the original number, and must be less than 10 times the original number. It cannot be 99, 999, etc, times the original number. We could stay with 9 times the original number, but I'm too lazy to solve it. ----- 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
Dear Richard: It is indeed a good idea to collect Murray Klamkin problems. I am particularly fond of one his problems-but it did not appear in Math journals, but in the book by L.A.Graham : Ingenious Mathematical problems and methods, Dover. Murray's problem,which appears as #72 in the book,reads: find the smallest number such that if the last digit is removed and placed at the beginning to become the first digit, this new number is nine times the original one. The solution given in the book itself is the astronomical fijure
10,112,359,550,561,797,752,808,988,764,044,943,820,224,719. 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