21) A number is first doubled and then decreased by one unit. Applying this procedure 2000 times in a row results in $2^{2001}+1$. What is the starting number?
(A) 1 (B) 2
(C) 3
(D) 4 (E) 5.
21) The answer is (C). Let $n$ be the starting number. At the first step, we get $2 n-1$, at the second step $4 n-3$, at the third step $8 n-7$, and so on. Therefore, it is possible to show by induction that at the $k$-th step we obtain $2^{k} n-2^{k}+1$, that is, $2^{k}(n-1)+1$. From the relation $2^{2000}(n-1)+1=2^{2001}+1$ it follows that $n=3$.

## SECOND SOLUTION

The inverse process to the one described is to increase the number by one unit and then halve it. Thus, starting from a number of the form $2^{n}+1$, we get $\frac{1}{2}\left(2^{n}+2\right)=2^{n-1}+1$. Starting from $2^{2001}+1$, after 2000 steps we arrive at $2^{1}+1=3$.