If we let
f(x) = (k1 x-r1)(k2 x-r2)...(kn x-rn)
then (r1/k1)(r2/k2)...(rn/kn) = 1024/1
Now, 1024 = 2^10, so all the k's are 1, and all the r's multiplied together are 2^10
the possible distinct roots are
2,4,8,16
so, f(x) = (x-2)(x-4)(x-8)(x-16) + 1024
has 4 values of x such that f(x) = 1024.
I'd say 3 is the max d such that there are n>d places where f(x) = 1024
If I'm way off base here, let me know. I'd be interested in how it's supposed to be done. What are you studying in the class now?