Given a polynomial \(f: \mathbb{Z}\to \mathbb{Z}\) with positive integer coefficients, how many evaluations of \(f\) does it take to obtain the polynomial?

(An \(f: \mathbb{R}\to \mathbb{R}\) polynomial with real coefficients would take the number of degrees plus 1 to specify, which, if it held in this case, would render the answer unbounded. But the correct answer in this case is surprisingly small.)

It takes only 2 evaluations. Suppose in the following that \(b>0\). Let us note that a polynomial \(f(b) = a_n b^n + … + a_0 b^0\) specifies essentially a base \(b\) representation of the number \(f(b)\), in that \(a_n a_{n-1} … a_0\) is an expansion of \(f(b)\) in base \(b\). The only problem is this expansion is non-unique, as it is possible for any \(a_j \ge b\).

However, it is not possible for any \(a_j \ge f(b) + 1\), since for all \(j\), \(f(b) \ge a_j\) by the problem statement and assumption on \(b\). Then take \(B = f(b) + 1\). Now we are guaranteed that \(a_n a_{n-1} … a_0\) is the unique (and canonical) base \(B\) expansion of \(f(B)\), from which the polynomial coefficients immediately obtain.

So the two evaluations are at \(f(b)\) and \(f(B=f(b)+1)\).

Example: \(f(b) = 3b^2 + 2b + 1\). Evaluate at, e.g., \(b=1\) to get \(f(1) = 6\). Then evaluate at \(B=f(1)+1=7\) to get \(f(7)=162=321_{7}\).

Also, world’s shortest escalator:

Say there are \(n\) cards up in the full deck. If you just divide the deck without flipping cards, then however you divide, one subdeck will have \(m\in [0,n]\) cards up, and the other subdeck will have the complement \(n-m\) cards up. So take \(n\) cards out of the original deck as subdeck A, and flip them over. If this subdeck A had \(m\) cards up originally, now it has \(n-m\) cards up. In the remaining cards forming subdeck B, there were already \(n-m\) cards up. So the two subdecks have an equal number of cards up.

As for the answer:

From left Japan’s Katsuya Okada, Korea’s Yu Myung-hwan, and China’s Yang Jiechi

I must say these are typical faces.