8. The school stage of the Magic and Wizardry Olympiad consists of 5 spells. Out of 100 young wizards who participated in the competition,

- 95 correctly performed the 1st spell
- 75 correctly performed the 2nd spell
- 97 correctly performed the 3rd spell
- 95 correctly performed the 4th spell
- 96 correctly performed the 5th spell.

What is the minimum number of students who could have correctly performed exactly 4 out of 5 spells under the described conditions?
# Answer: 8.

Solution. The number of students who correctly performed all spells is no more than 75, since only 75 students correctly performed the second spell. The number of students who made mistakes in the 1st, 3rd, 4th, or 5th spells is no more than $(100-95)+(100-97)+(100-95)+(100-96)=$ $5+3+5+4=17$. If a student made a mistake in at least two spells, then they definitely made a mistake in a spell other than the second. Therefore, the number of students who made a mistake in at least two spells does not exceed 17. Then the desired number of students is no less than $100-75-17=8$.

It remains to show that such a number of students is possible. Indeed, let the first 25 students make a mistake in the second spell. Among them, five made a mistake in the first, three in the third, five in the fourth, and four in the fifth. Since $5+3+5+4=17$ is less than 25, these 17 students can be different.

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