Let a = 1.
p(1), which is the sum of the coefficients,
is clearly greater than or equal to each of
the coefficients.
Let b = 10^k > p(1) for some k.
You can read off the coefficients of p(x) in
the answer to p(b) in groups of k digits.
Note: Any b > max(1,p(1)) will work, but you
will then have to convert p(b) to base b. This
way you can perform the stunt over and over
again without your opponent figuring out how
you did it.
|
