1. Several numbers were written on the board, their arithmetic mean was equal to $M$. They added the number 15, after which the arithmetic mean increased to $M+2$. After that, they added the number 1, and the arithmetic mean decreased to $M+1$. How many numbers were on the board initially? (Find all options and prove that there are no others.)
Answer: 4.

Let there be $k$ numbers in the original list with a sum of $S$. Then, by the condition,

$$
\frac{S+15}{k+1}-\frac{S}{k}=2, \quad \frac{S+15}{k+1}-\frac{S+16}{k+2}=1.
$$

By bringing to a common denominator and transforming in an obvious way, we get that these equations are equivalent to the following two:

$$
15 k-S=2 k(k+1), \quad S-k+14=(k+1)(k+2)
$$

Adding them, we arrive at the equation for $k$:

$$
14(k+1)=(k+1)(2 k+k+2),
$$

from which $k=4$.

 GHz The correct system of equations is written, but solved incorrectly. 3 points

- One of the equations is obtained correctly, the second one incorrectly. 2 points
- The correct answer without justification. (An example of a list of four numbers satisfying the condition is not considered a justification.) 1 point