91 (1201). Find the smallest integer that needs to be added to the expression $(a+2)(a+5)(a+8)(a+11)$ so that the resulting sum is positive for any value of $a$.
Solution. Let $b$ be the smallest integer such that the expression $(a+2)(a+5)(a+8)(a+11)+b$ is positive for any value of $a$. Transform the obtained expression into a quadratic trinomial:

$$
\begin{gathered}
(a+2)(a+5)(a+8)(a+11)+b= \\
=(a+2)(a+11)(a+5)(a+8)+b= \\
=\left(a^{2}+13 a+22\right)\left(a^{2}+13 a+40\right)+b= \\
=c(c+18)+b=c^{2}+18 c+b, \text { where } c=a^{2}+13 a+22 .
\end{gathered}
$$

Find the discriminant of the trinomial $c^{2}+18 c+b: D=324-4 b$. Since the quadratic trinomial $c^{2}+18 c+b$, which has a positive leading coefficient, takes positive values for any value of the variable only if its discriminant is negative, we require that the inequality $324-4 b < 0$ holds, i.e., $b > 81$.

The smallest integer satisfying the condition $b > 81$ is 82. Therefore, $b=82$ is the smallest integer for which the trinomial $c^{2}+18 c+b$ takes positive values for any $c$, and thus the expression $(a+2)(a+5)(a+8)(a+11)+82$ is positive for any value of $a$.

The solution to the problem can be simplified if we choose $c$ to be the expression $a^{2}+13 a+31$. Then $\left(a^{2}+13 a+22\right) \times$ $\times\left(a^{2}+13 a+40\right)+b=(c-9)(c+9)+b=c^{2}-81+b>0, \quad$ from which $b>81$.