   SEARCH HOME Math Central Quandaries & Queries  Question from Jash, a student: Assume there is a number system of base m. The one property of this system is: If 2 numbers written in this system, which have 'a' and 'b' as the number of digits are multiplied, then the product of the 2 numbers will have a number of digits which is a function f(a,b). In other words, as long as the number of digits of the 2 numbers are constant, the number of digits of their product is a constant. Find m. Jash,
This is not possible [if we assume the base of a number system is a whole number >1.]

The proof is not hard to find; consider the smallest number with $n$ digits $m^{n-1}$, and the largest $m^n - 1.$

Show that the quotient of their squares

$\frac{(m^n - 1)^2}{(m^{n-1})^2}$

is always strictly bigger than $m.$ Deduce that $(m^n - 1)^2$ always has exactly one more digit than $(m^{n-1})^{2}.$

Good Hunting!
RD     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.