1. **Problem statement:** We are given a set of 9 children participating in 3 tasks. 5 children solved the first task, 6 solved the second, and 7 solved the third. Every child solved at least one task. We want to find how many children could have solved all three tasks. 2. **Formula and rules:** Use the principle of inclusion-exclusion for three sets $A$, $B$, and $C$ representing children who solved tasks 1, 2, and 3 respectively. The formula for the union of three sets is: $$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |A \cap C| + |A \cap B \cap C|$$ Since all children solved at least one task, $|A \cup B \cup C| = 9$. 3. **Known values:** $$|A|=5, \quad |B|=6, \quad |C|=7, \quad |A \cup B \cup C|=9$$ 4. **Rearranging the formula to find $|A \cap B \cap C|$:** $$|A \cap B \cap C| = |A \cup B \cup C| - (|A| + |B| + |C|) + (|A \cap B| + |B \cap C| + |A \cap C|)$$ 5. **Maximizing $|A \cap B \cap C|$:** To find the maximum number of children who solved all three tasks, minimize the pairwise intersections $|A \cap B|$, $|B \cap C|$, and $|A \cap C|$. 6. **Minimum pairwise intersections:** Since $|A|=5$, $|B|=6$, and $|C|=7$, the minimum pairwise intersections can be estimated by: $$|A \cap B| \geq |A| + |B| - 9 = 5 + 6 - 9 = 2$$ $$|B \cap C| \geq 6 + 7 - 9 = 4$$ $$|A \cap C| \geq 5 + 7 - 9 = 3$$ 7. **Calculate $|A \cap B \cap C|$ with minimum pairwise intersections:** $$|A \cap B \cap C| = 9 - (5 + 6 + 7) + (2 + 4 + 3) = 9 - 18 + 9 = 0$$ 8. **Maximize triple intersection:** If pairwise intersections are larger, triple intersection can be larger. The maximum triple intersection is limited by the smallest set size, which is 5. 9. **Conclusion:** The number of children who solved all three tasks could be as high as 5. **Final answer:** 5