5. Out of two hundred ninth-grade students, 80% received excellent grades on the first exam, 70% on the second exam, and 59% on the third exam. What is the smallest number of participants who could have received excellent grades on all three exams?

Answer: 18.
Let $M_{i}$ be the number of students who received excellent grades only on the $i$-th exam; $M_{ij}$ be the number of students who received excellent grades only on exams $i$ and $j$; $M_{123}$ be the number of students who received excellent grades on all three exams. Then, according to the problem,

$$
\left\{\begin{array}{l}
M_{1}+M_{12}+M_{13}+M_{123}=160 \\
M_{2}+M_{12}+M_{23}+M_{123}=140 \\
M_{3}+M_{23}+M_{13}+M_{123}=118
\end{array}\right.
$$

Adding all three equations:

$$
M_{1}+M_{2}+M_{3}+2\left(M_{12}+M_{13}+M_{23}\right)+3 M_{123}=418
$$

Notice that

$$
M_{1}+M_{2}+M_{3}+M_{12}+M_{13}+M_{23}+M_{123} \leq 200
$$

(on the left side of this inequality, the number of students who received at least one excellent grade is recorded, and on the right side - the total number of students). Subtracting (2) from (1), we get

$$
M_{12}+M_{13}+M_{23}+2 M_{123} \geq 218 \Leftrightarrow\left(M_{12}+M_{13}+M_{23}\right) \geq 218-2 M_{123}
$$

Since $M_{23} \leq 200-160=40, M_{13} \leq 200-140=60, M_{12} \leq 200-118=82$, then $M_{12}+M_{13}+M_{23} \leq 182$. Considering (3), we get that $182 \geq 218-2 M_{123}$, from which $M_{123} \geq 18$.

It is not difficult to verify that the value $M_{123}=18$ can be achieved. For this, take $M_{12}=82$, $M_{13}=60, M_{23}=40, M_{1}=M_{2}=M_{3}=0$.