24) Enrico's powerful computer was recently infected by a virus that gradually erases all memory. Every minute, the erased region is quadrupled and then increased by 1 byte. Enrico freezes the computer when the situation is already compromised: he calculates that not only will the virus corrupt all the memory in one minute, but also that it would suffice to triple the erased region and add 1 byte to exactly exhaust the computer's memory, which amounts to $2.5 \mathrm{~Gb}=2.5 \cdot 2^{30}$ bytes. How large was the erased region 14 minutes ago?
(A) 1 byte
(B) 2 bytes
(C) 3 bytes
(D) 4 bytes
(E) 16 bytes.
24) The answer is (C). If $n$ denotes the number of compromised bytes at a certain moment, after one minute these become $4 n+1$, then $4^{2} n+4+1$, then $4^{3} n+4^{2}+4+1$, and so on, until after 14 minutes they are

$$
4^{14} n+4^{13}+\cdots+4+1=n \cdot 2^{28}+2^{26}+\cdots+2^{2}+1
$$

Double this number is $2 n \cdot 2^{28}+2^{27}+\cdots+2^{3}+2$, so the triple (by adding) is

$$
3 n \cdot 2^{28}+2^{27}+2^{26}+\cdots+2+1=3 n \cdot 2^{28}+2^{28}-1
$$

Equating the last value with the total memory of the PC minus one byte, $2.5 \cdot 2^{30}-1=10 \cdot 2^{28}-1$, we find that $3 n+1=10$ and thus $n=3$.

To convince oneself that $2^{27}+2^{26}+\cdots+2+1=2^{28}-1$, one can use the notable product $x^{28}-1=$ $(x-1)\left(x^{27}+x^{26}+\cdots+x+1\right)$ with $x=2$, or add 1 to both members, noting that

$$
\begin{aligned}
1+1+2+2^{2}+2^{3} \cdots+2^{27} & =2+2+2^{2}+2^{3}+\cdots+2^{27} \\
& =2^{2}+2^{2}+2^{3}+\cdots+2^{27}=\cdots=2^{27}+2^{27}=2^{28}
\end{aligned}
$$