Selection Tests of the Belarusian Team to the IMO

If the polynomial $P(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ is expandable, then it can be represented as
$$P(x) = (x^m + b_{m-1}x^{m-1} + \cdots + b_1x + b_0) \cdot (x^k + c_{k-1}x^{k-1} + \cdots + c_1x + c_0),$$
where $m \ge k \ge 1$, $m + k = n$ and numbers $b_i, c_i$ are all non-negative integers. Since the coefficients of the polynomial $P(x)$ are not greater than $100$, all these numbers do not exceed $100$ and, moreover, $b_i \cdot c_i \le 100$ for any $i$ from $0$ to $k-1$. This means that for a fixed $i$ from $0$ to $k-1$ the number of possible pairs $(b_i, c_i)$ does not exceed
$$2 \cdot 11 \cdot 101 < \frac{101^2}{4},$$
because at least one of $b_i$ and $c_i$ is not greater than $\sqrt{100} = 10$.
Thus, the numbers $b_{m-1}, b_{m-2}, \dots, b_k$ can be chosen in no more than $101^{m-k}$ ways, and the numbers $b_0, c_0, b_1, c_1, \dots, b_{k-1}, c_{k-1}$ you can select no more than $\left(\frac{101^2}{4}\right)^k$ ways. We find that the number of expandable polynomials can be estimated from above by the expression
$$\sum_{k=1}^{[n/2]} 101^{m-k} \left(\frac{101^2}{4}\right)^k = 101^n \sum_{k=1}^{[n/2]} \frac{1}{4^k} < 101^n \frac{1/4}{1 - 1/4} = \frac{1}{3} \cdot 101^n$$

Since the number of polynomials $P(x) = x^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$ with non-negative integer coefficients not exceeding $100$ is equal to $101^n$, then there are more than $\frac{2}{3} \cdot 101^n$ non-expandable polynomials, that is, at least twice as many as expandable ones.