1. 25 of the participants solve at least one problem each.

25 = a + b + c + ab + ac + bc + abc

2. Of all those who don’t solve problem A, the number who solve B is twice the number who solve C.

b + bc = 2*(c + bc) => bc = b - 2c

3. The number of participants who solve only problem A is one more than the number of those who solve A and at least one other problem.

a = 1 + ab + ac + abc => ab + ac + abc = a - 1

4. Of all participants who solve just one problem, half do not solve problem A.

b + c = 1/2 * (a + b + c) => 1/2 * b + 1/2 * c = 1/2 * a => a = b + c

If we substitute 2, 3, and 4 into 1, we are left with an equation in b and c:

2,3 => 1: a + b + c + a - 1 + b - 2c = 25 => 2a + 2b - c = 26

4 => 1: 2b + 2c + 2b - c = 26 => c = 26 - 4b

As both b and c must be positive integers, here are all possible combinations:

b c

1 22

2 18

3 14

4 10

5 6

6 2

But from 2 we know that b = bc + 2c

Therefore, b must be at least equal to 2c. From the table above, it is clear that this is only possible when b is 6.

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