Harvard-MIT Mathematics Tournament

Solution:

Answer: $86400000$

For each nonnegative integer $i$, let $x^{\underline{i}} = x(x-1) \cdots (x-i+1)$. (Define $x^{0} = 1$.)

Lemma: Each polynomial with integer coefficients $f$ can be uniquely written in the form
$$f(x) = a_{n} x^{\underline{n}} + \ldots + a_{1} x^{\underline{1}} + a_{0} x^{\underline{0}}, \quad a_{n} \neq 0$$
Proof: Induct on the degree. The base case (degree $0$) is clear. If $f$ has degree $m$ with leading coefficient $c$, then by matching leading coefficients we must have $m = n$ and $a_{n} = c$. By the induction hypothesis, $f(x) - c x^{\underline{n}}$ can be uniquely written as $a_{n-1} x^{\underline{n-1}} + \ldots + a_{1} x^{\underline{1}} + a_{0} x^{\underline{0}}$.

There are $120$ possible choices for $a_{0}$, namely any integer in $[0,120)$. Once $a_{0}, \ldots, a_{i-1}$ have been chosen so $0 \leq P(0), \ldots, P(i-1) < 120$, for some $0 \leq i \leq 5$, then we have
$$P(i) = a_{i} i! + a_{i-1} i^{\underline{i-1}} + \cdots + a_{0}$$
so by choosing $a_{i}$ we can make $P(i)$ any number congruent to $a_{i-1} i^{\underline{i-1}} + \cdots + a_{0}$ modulo $i!$. Thus there are $\frac{120}{i!}$ choices for $a_{i}$. Note the choice of $a_{i}$ does not affect the value of $P(0), \ldots, P(i-1)$. Thus all polynomials we obtain in this way are valid. The answer is
$$\prod_{i=0}^{5} \frac{120}{i!} = 86400000$$

Solution:
For each nonnegative integer $i$, let $x^{\underline{i}} = x(x-1) \cdots (x-i+1)$. (Define $x^{0} = 1$.)

Lemma: Each polynomial with integer coefficients $f$ can be uniquely written in the form
$$f(x) = a_{n} x^{\underline{n}} + \ldots + a_{1} x^{\underline{1}} + a_{0} x^{\underline{0}}, \quad a_{n} \neq 0$$
Proof: Induct on the degree. The base case (degree $0$) is clear. If $f$ has degree $m$ with leading coefficient $c$, then by matching leading coefficients we must have $m = n$ and $a_{n} = c$. By the induction hypothesis, $f(x) - c x^{\underline{n}}$ can be uniquely written as $a_{n-1} x^{\underline{n-1}} + \ldots + a_{1} x^{\underline{1}} + a_{0} x^{\underline{0}}$.

There are $120$ possible choices for $a_{0}$, namely any integer in $[0,120)$. Once $a_{0}, \ldots, a_{i-1}$ have been chosen so $0 \leq P(0), \ldots, P(i-1) < 120$, for some $0 \leq i \leq 5$, then we have
$$P(i) = a_{i} i! + a_{i-1} i^{\underline{i-1}} + \cdots + a_{0}$$
so by choosing $a_{i}$ we can make $P(i)$ any number congruent to $a_{i-1} i^{\underline{i-1}} + \cdots + a_{0}$ modulo $i!$. Thus there are $\frac{120}{i!}$ choices for $a_{i}$. Note the choice of $a_{i}$ does not affect the value of $P(0), \ldots, P(i-1)$. Thus all polynomials we obtain in this way are valid. The answer is
$$\prod_{i=0}^{5} \frac{120}{i!} = 86400000$$