b) The numbers $1,2^{1}, 2^{2}, 2^{3}, \ldots, 2^{10}$ are written on the board. It is allowed to erase any two numbers and write down their difference - a non-negative number. After several such operations, only one number will remain on the board. What can this number be?
a) 97 can be obtained, for example, as follows. By successively subtracting from 16 the numbers 8, 4, 2, 1, we get 1. On the board, the numbers 1, 32, 64, 128 remain. Next: $64 - 32 = 32$, $32 - 1 = 31$, $128 - 31 = 97$.

b) We will prove that if the numbers $1, 2, \ldots, 2^n$ are written on the board, then after $n$ operations described in the condition, any odd number from 1 to $2^n - 1$ can be obtained. It is obvious that numbers greater than $2^n$ do not appear on the board. It is also easy to see that there is always exactly one odd number on the board. Therefore, the last remaining number on the board is odd. We will prove by induction on $n$ that all the specified numbers can be constructed. Base case. With the numbers 1 and 2, only the number 1 can be obtained.

Inductive step. Suppose the numbers $1, 2, \ldots, 2^{n+1}$ are written on the board. Any odd number less than $2^n$ can be obtained in $n+1$ operations (on the first step, we erase $2^{n+1}$ and $2^n$ and write $2^n$, then proceed by the inductive hypothesis). Odd numbers from $2^n + 1$ to $2^{n+1} - 1$ can be written in the form $2^{n+1} - a$, where the number $a$ can be obtained from the set $1, 2, \ldots, 2^n$. On the last step, we subtract $a$ from $2^{n+1}$.

## Answer

a) It can; b) any odd number from 1 to $2^{10} - 1$.