9.2. Represent the number 100 as the sum of the maximum possible number of pairwise coprime natural numbers. Explanation: the condition means that the greatest common divisor of any two numbers used in the sum is 1.
Answer: 100 is the sum of all nine of the first prime numbers from 2 to 23 inclusive.

Solution. Indeed, $100=2+3+5+7+11+13+17+19+23$, the number of addends is 9, which is the maximum possible. If 100 were the sum of not less than ten different pairwise coprime natural numbers, then due to their pairwise coprimality, the prime factors of any two of them would be different. By leaving any prime divisor from each number not equal to 1, we would obtain ten or more different numbers, each of which is prime except, possibly, one equal to 1, the sum of which does not exceed 100. The latter is impossible, since the sum of the ten smallest prime numbers is $2+3+5+7+11+13+17+19+23+29=129>100$, and the sum of 1 and the nine smallest prime numbers is $1+2+3+5+7+11+13+17+19+23=101>100$.

Grading criteria. The example $2+3+5+7+11+13+17+19+23=100$ is provided: 3 points. The maximality of the example is proven: 4 points. Total 7 points.

If the case where one of the numbers is equal to one is not considered: minus 2 points. All references to the example as the "best case" in attempts to justify maximality: 0 points for justification.