1. Given $n>2$ natural numbers, among which there are no three equal, and the sum of any two of them is a prime number. What is the largest possible value of $n$?
1. Answer: 3. Note that the triplet $1,1,2$ satisfies the condition. Suppose there are more than 3 numbers. Consider any 4 of them a, b, c, d. Among these four numbers, there cannot be two even numbers, otherwise their sum would be an even prime greater than two. Therefore, at least 3 of them must be odd. Their pairwise sums are even primes. Then all three considered numbers must be 1, which violates the condition.

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only answer - 0 points

example for three numbers provided - 2 points.