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| Math Central | Quandaries & Queries |
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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
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