1. The arithmetic mean of several distinct prime numbers is 27. Determine the largest prime number among them.
1. Let $P$ be the set of prime numbers under investigation, and first show that $2 \notin P$. The number 2 is the only prime number that is not odd. If $2 \in P$, the sum of all primes in $P$ (an odd number of them) would be even, and the sum of an even number of them would be odd, so the arithmetic mean could not be the odd number 27. Therefore, $2 \notin P$.

Since the number 27 is not a prime number, the set $P$ is not a singleton, and for its largest element $p^{*}$, it holds that $p^{*} > 27$. Now, we will use this obvious fact: The arithmetic mean $A$ of a set of real numbers decreases whenever we add a number smaller than $A$ to the set or remove a number larger than $A$ from the set. Therefore, let's add all the missing prime numbers $p, 2 < p < 27$ to the set $P$ and remove all prime numbers $p, 27 < p < p^{*}$ (if any exist). This gives us a set of nine prime numbers $\left\{3,5,7,11,13,17,19,23, p^{*}\right\}$, for which the arithmetic mean (which may no longer be an integer) satisfies the estimate

$$
\frac{3+5+7+11+13+17+19+23+p^{*}}{9} \leqq 27
$$

(Equality holds if we neither added any prime number nor removed any.) From this, we get $p^{*} \leqq 145$. The largest prime number that satisfies this inequality is 139.

The value $p^{*}=139$ is possible, as shown by the example

$$
P=\{3,5,7,11,13,17,19,29,139\},
$$

which we find when we replace the prime number 23 in the sum $3+5+7+11+13+17+19+23$ with a prime number that is $145-139=6$ larger. (If we had not realized that $2 \notin P$ beforehand, we would have obtained from the inequality

$$
\frac{2+3+5+7+11+13+17+19+23+p^{*}}{10} \leqq 27
$$

the weaker estimate $p^{*} \leqq 170$. Then it would be necessary to rule out the values $p^{*}=167,163,157$, 151,149. In doing so, we would likely realize why $2 \notin P$.)

For $p^{*}=139$, there is one more set $P$ with the required properties. It is an eleven-element set

$$
P=\{3,5,7,11,13,17,19,23,29,31,139\}
$$

For a complete solution, 6 points are awarded. For the reasoning leading to the exclusion of 2, 2 points are awarded, and for providing an example for $p^{*}=139$, 2 points are awarded.