9.5. Given 10 numbers: $10, 20, 30, \ldots, 100$. With them, you can perform the following operation: choose any three and add one to each of the chosen numbers. The same operation is then performed on the resulting 10 numbers, and so on. Can you, after several operations, obtain: a) all the same numbers? b) all numbers equal to 200?
Answer: a) yes, b) no. Solution. a) We will show how to perform three operations on the numbers $a_{1}, a_{2}, \ldots, a_{10}$ so that after several sets of three operations, all numbers become equal. Let $m$ be the smallest and $M$ be the largest of the numbers $a_{1}, a_{2}, \ldots, a_{10}$. We calculate the sum

$$
S=\left(a_{1}-m\right)+\left(a_{2}-m\right)+\ldots+\left(a_{10}-m\right)
$$

If $M=m$, then all numbers are the same and $S=0$. If $M>m$, then $S>0$ and we will perform the following three operations, which will reduce $S$ by one, and thus after several such operations, $S$ will become zero. These three operations consist of adding 1 to three triplets of numbers; if, for definiteness, $a_{10}=M$, then these triplets are: $\left\{a_{1}, a_{2}, a_{3}\right\},\left\{a_{4}, a_{5}, a_{6}\right\},\left\{a_{7}, a_{8}, a_{9}\right\}$. As a result of these three operations, all numbers except $a_{10}$ will increase by 1 and $m$ will increase by 1, i.e., $S$ will decrease by 1. b) We will show that this is impossible. After each operation, the sum of the numbers increases by 3, so after $n$ operations, it will increase by $3n$. Initially, this sum was $10+20+\ldots+100=550$, i.e., it gave a remainder of 1 when divided by 3. At the end, the sum should equal 2000, i.e., the remainder when divided by 3 should become 2, which is impossible.