3. On the hundredth year of his reign, the Immortal Treasurer decided to start issuing new coins. In this year, he put into circulation an unlimited supply of coins with a value of $2^{100}-1$, the following year - with a value of $2^{101}-1$, and so on. As soon as the value of the next new coin can be exactly matched using the previously issued new coins, the Treasurer will be removed from office. In which year of his reign will this happen? (I. Bogdanov)
Answer. On the two hundredth. Solution. Let on the $k$-th year of reign $2^{k}-1$ can be made up of previously issued coins: $2^{k}-1=a_{1}+\ldots+a_{n}=N-n$, where $N$ is the sum of powers of two, each of which is divisible by $2^{100}$. Since $2^{k}$ is also divisible by $2^{100}$, the number $n-1$ must be divisible by $2^{100}$. Since, obviously, $n>1$, we get $n \geq 2^{100}+1$, from which $2^{k}-1 \geq\left(2^{100}-1\right)\left(2^{100}+1\right) \geq 2^{200}-1$, that is, $k \geq 200$, and the Treasurer will not be removed earlier than the 200th year. But on the 200th year, he will be removed, since $2^{200}-1=\left(2^{100}+1\right)\left(2^{100}-1\right)$.