... From: Jose Ramón Brox If they exist, you have several solutions, since your number can have 3 digits in several bases x. [...] ...

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, Oct 11, 2005

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----- Original Message -----
From: "Jose Ramón Brox" <ambroxius@...>

If they exist, you have several solutions, since your number can have 3 digits in several
bases x. [...]

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There is always at least one base in which N>2 has 3 digits:

if x = N+1 then N = N)x has 1 digit
if x = N then N = 10)x has 2 digits
if x = Floor(Sqrt(N)) then N)x has 3 digits, since
x^2 = [Floor(Sqrt(N))]^2 < Sqrt(N)^2 = N
and (Floor(Sqrt(N)))^2 >= (Sqrt(N)-1)^2 = N - 2Sqrt(N) +1 -->
--> 2*x^2 >= 2N - 4Sqrt(N) +1 > N (if N>2)

So we have that x^2 < N < 2*x^2 and then, if x>2, N has 3 digits in base x =
Floor(Sqrt(N)).

Examples:

N = 1532 --> x = 39 --> N)x = 10B
N = 54489 --> x = 233 --> N)x = 10(200) (I don't have a proper symbol for the digit
(200) in base 233)
N = 9999 --> x = 99 --> N)x = 120

So here you have a procedure to find ONE way of expressing N as a number of 3 digits, with
a = 1: take x = Floor(Sqrt(N)), then compute R = mod(N/x^2) and compare it with x; if R <
x, then your number N is 10R)x. If R >= x, then divide again to get S = mod(R/x) and
finally your number is 1RS)x.

Regards. Jose Brox

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