A big power

Thank you to Louise, Niharika from Rugby School in England, JuHo from Beijing BISS International School in Korea and Tomasso from the British School of Milan in Italy for submitting solutions to this problem.

Here is Louise's answer to the first part:

We want to estimate $2^{2000}$ as a power of $10$ so we are trying to solve $2^{2000}=10^x$ for $x$. The first step is to take logs of both side with respect to a suitable basis such as $2$, so we get $2000=x\log_{2}{10}$. Now all that remains is to estimate $\log_{2}{10}$. We observe that it lies between $3$ and $4$ since $2^3=8$ and $2^4=16$. Hence as a rough approximation $x$ lies between $500$ and $700$, so $2^{2000}$ lies between $10^{500}$ and $10^{700}$.

JuHo has managed to improve on this range for $x$:

$2^{2000}=10^x$

$x=2000 \log_{10}{2}$

We can get two inequalitites for $\log_{10}{2}$.

Firstly, $\log{8}<\log{10}$, so $3\log{2}<1$ so $\log_{10}{2}<\frac{1}{3}=0.\dot{3}$.

Secondly $\log{1024}>\log{1000}$ so $10\log{2}>3$, so $\log_{10}{2}>\frac{3}{10}=0.3$.

In summary $0.3 < \log_{10}{2} < 0.\dot{3}$,

so $600

< x < 667$

JuHo found the first 5 digits of $2^{2000}$ by calculating it to log base 10 first:

$x=2^{2000}$

$\log_{10}{x}=2000\log_{10}{2}=602.059913\dots$

$x=10^{602.069913\dots}=10^{602} \times 10^{0.059913\dots}=1.14813\dots \times 10^{602}$

This shows that the first 5 digits of $x=2^{2000}$ are $1,1,4,8,1$ in that order.

Niharika noticed a neat way to find the final digit of $2^{2000}$ using the given fact that $2^{16}=65536$:

The first thing that is helpful to notice is that since $2000=16 \times 125$, $2^{2000}=(2^{16})^{125}=65536^{125}$

Hence, since $2^{2000}$ is a power of something ending in $6$, it will also end in $6$ because $6^2=36$

For finding the last two digits of $2^{2000}$, both Tomasso and JuHo had the following idea of trying to look for a pattern in the sequence of the last two digits of powers of 2.

The sequence whose $n^{th}$ term is the last two digits of $2^n$ is:

$02,04,08,16,32,64,28,56,12,24,48,96,92,84,68,36,72,44,88,76,52,04,08,16,32,64,...$

We see that the $22^{nd}$ term is the same as the $2^{nd}$ term and after this the sequence will repeat the same 20 terms over and over.

$2^{2000}$ corresponds to the $2000^{th}$ term of this sequence. The remainder when $2000-1$ is divided by $20$ is $19$, and hence, because the recurring starts from the second term, the last two digits of $2^{2000}$ is the same as the $20^{th}$ term of the sequence, so the last two digits are $76$.

Niharika found the last two digits by a similar method but considered powers of $36$ instead of $2$ as she used again the fact that $2^{2000}= 65536^{125}$.