Each of x₁, x₂ and x₃ corresponds to a nonzero digit, whether same or different, in positive integer base b.

Determine all possible value(s) of b ≤ 100 such that this equation has at least one valid solution:

x₁x₂x₃ = x₁^x₁ + x₂^x₂ + x₃^x₃

Note: x₁x₂x₃ corresponds to concatenation of the three digits.

given x=x1 y=x2 and z=x3 then for base b we want

x1*b^2+x2*b+x3=x1^x1+x2^x2+x3^x3

let n=x1^x1+x2^x2+x3^x3

then solving for b we get

b=(Sqrt[y^2+4*x*(n-z)]-y)/(2x) (1)

so for any given x,y,z we want the required b from (1) to both be an integer and be strictly greater than each of x,y,z.  Using this I did a search for x,y,z between 1 and 99 and found 2 solutions beyond the 4 already found and they are

Base 904: 171

Base 215: 126

now looking at these and the other 4 it appears that there may not be any solutions with x,y,z>7 but I have yet to work out a proof of this.

Posted by Daniel on 2009-04-13 14:54:01