4 ms·

[redacted because I was completely wrong, missing the obvious ;-)] ... See child comment.

• ColinWright 8y ago
  > No it is not. Your statement of the problem does not rule out only the coefficient of x^0 being positive, and the rest being zero. You are correct that we permit any and/or all coefficients to be zero, allowing the case that p(x)=665. That is intentional. > Just saying the coefficients are all nonnegative, the polynomial p(k) = 665 also is a solution for the case k = 6 and p(6) = 665. No it isn't, because you have evaluated at 6, which is not larger than all the coefficients. Specifically, 6 is not larger than 665. So I still claim that the question is fine as it stands. I do appreciate you trying to poke holes in it, though. Thanks. (Added in edit: Given that you've retracted that, would you like me to remove this comment?)