In a mathematics competition, 30 students participated. Among them,

the 1st problem was solved by 20,

the 2nd problem was solved by 16,

the 3rd problem was solved by 10,

the 1st and 2nd problems were solved by 11,

the 1st and 3rd problems were solved by 7,

the 2nd and 3rd problems were solved by 5,

all three problems were solved by 4.

students. How many students participated in the competition without solving a single problem?
I. solution: Thirty students participated in the competition. If each solver had solved only one problem, then the number of students who did not solve the problem ($n$) would be obtained by subtracting the sum of the number of solvers of the 1st, 2nd, and 3rd problems from the total number of participants. In our case, this value is

$$
30-(20+16+10)
$$

Now, taking into account those who solved two problems each, we see that they appear twice in the above formula's subtrahend, so we need to add the sum of their numbers (1) to the right side of the difference. The thus modified value is:

$$
30-(20+16+10)+(11+7+5)
$$

However, since students who solved all three problems also participated, we have considered these three times in the last parenthesis of expression (2), whereas according to the previous, they should only be considered twice. Therefore, our formula will now be correct if we subtract the number of these from the right side.

Thus, at the mathematics competition,

$$
n=30-(20+16+10)+(11+7+5)-4=3
$$

students did not solve a single problem.

# II. solution: 

Solved all three problems........................... 4 students

solved only the 1st and 2nd problems $11-4=7$ students

" the 1st and 3rd " " " $7-4=3$

" the 2nd and 3rd ", " " $5-4=1$

" the 1st " " ", $20-(4+7+3)=6 \quad "$

" the 2nd " " " $16-(4+7+1)=4 \quad "$

" the 2nd " " " $10-(4+3+1)=2 \quad "$

Solved at least one problem ........................... 27 students, so $30-27=3$ students did not solve a single problem.