Without using any arithmetical symbols (+, , *, /, or similar; other math symbols; decimal comma or periods; letters; even parentheses) or, in short, anything but the digits, build a number with the digits 1, 3, 5, 7 and 9, that is equal to a number built with the digits 2, 4, 6 and 8 (each digit used once and only once).
Note: This is not a trick. It was extracted from a book edited by Angela Dunn, a mathematician who gathered problems that appeared in many scientific periodical revues!
I agree with Richard, using symbols in the digits place to denote values greater than the base is well defined, but it would be the height of politeness to only call it unconventional. Thus, my first solution is unsatisfying. It occured to me that when using bases greater than nine, it is unclear what base is being used for the base. This leads me to a new solution that does not use the "big digit" trick:
31_5_7_9 = 24_6_8
Again, the underscore indicates to subscript. In words this is 31 base (5 base (7 base 9)) = 24 base (6 base 8). This isn't unique; I could use Ken Haley's idea and do:
31_5_7^9=24_6_8 or 31_5_9^7
What do you think?
Edited on July 11, 2005, 9:38 pm

Posted by owl
on 20050711 21:34:06 