Is there a power of 666 such that its decimal notation starts with the digits 123456789?
One may replace "starts" by "ends" for a much easier challenge ;-)
The second part is obviously impossible: all powers of 666 end in "6", so no one can end in "9".
For the first part, we need an integer K so 666^K= 123456789.something x 10^other, so the fractional part of K x log10(666) approximately equals that of log10(123456789). By
Kronecker's theorem, there exist infinite values of K such that satisfy this equation.
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