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This question was similarly answered at How many digits does $2^{1000}$ contain? . I saw in a book asking to do the same for $2^{100}$ without taking logaritms. There is a hit as following: " $2^{10}=1024$ and note that if $0<r<b<a$ then $\frac{a}{b}<\frac{a-r}{b-r}$ ". Honestly, I couldn't use this hint to solve the problem. Thanks.

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  • $\begingroup$ Which book was that? $\endgroup$
    – lhf
    May 26, 2012 at 2:10
  • $\begingroup$ which book? why wouldn't you provide a reference to this :/ $\endgroup$
    – baxx
    Jul 7, 2016 at 14:46

5 Answers 5

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Note that $2^{10} = 1024$ so $1000^{10} < 2^{100} = 1024^{10} < 1100^{10} = 1.1^{10}\cdot 1000^{10}$. We can estimate \begin{equation*} 1.1^{10} = 1.21^5 < 1.3^5 = 1.69^{2.5} < 2^3 = 8, \end{equation*} so $10^{30}<2^{100}<8\cdot 10^{30}$ and $2^{100}$ has 31 digits. On the other hand this doesn't use the given hint and the chain of equalities amounts to showing that $\log_{10}(1.1)<0.1$, so in some sense logarithms are involved.

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Write $2^{100}=(10^3+24)^{10}=10^{30}(1+0.024)^{10}$.

Consider $f(x)=x^{10}$. By the mean value theorem, $f(1+h)=f(1)+f'(\xi)h$, for some $\xi\in(1,1+h)$. Let $u=(1+h)^{10}$. Then $u=1+10\xi^9h<1+10(1+h)^9h=1+10\dfrac{u}{1+h}h$ and so $u<\dfrac{1+h}{1-9h}$.

When $h=0.024$, we get $u<\dfrac{1.024}{0.784}<10$, which implies that $10^{30}<2^{100}=10^{30}u<10^{31}$.

This means that $2^{100}$ has 31 digits. (By computing $\dfrac{1.024}{0.784}$, we get $2^{100}<1.31\cdot 10^{30}$.)

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Consider $\frac{2^{100}}{10^{30}} = \left( \frac{1024}{1000} \right)^{10}$.

Use your hint with $a=1024$, $b=1000$ and $r=500$, then $$\frac{1024}{1000} < \frac{524}{500} = 1 + \frac{6}{125}$$

Now, use $1+a < \mathrm{e}^a$,
$$\frac{2^{100}}{10^{30}} < \left(1+\frac{6}{125}\right)^{10} < \exp\left(\frac{60}{125}\right) < \sqrt{\mathrm{e}} < 2$$

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    $\begingroup$ It doesn't seem like the hint is buying you anything: if you skip it the $6$ changes to a $3$ and the rest of the argument still works. $\endgroup$
    – Noah Stein
    May 25, 2012 at 16:43
  • $\begingroup$ @NoahStein Quite true! $\endgroup$
    – Sasha
    May 25, 2012 at 16:46
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right now it is been some times that I not on maths rails but I will suggest it with a simple log function: $$ \log_{10}2^{100} = 100*log_{10}2 = 31 $$

where 10 is our decimal digit base. Maybe I am wrong?

Sorry Just saw not without taking logaritms

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  • $\begingroup$ OK. Your help was Welcome. :-) $\endgroup$
    – Mikasa
    May 25, 2012 at 20:54
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Answer should be $334$

By hand-multiplication to find different powers of $2$, starting from $1$ and continuing in this way by keeping on multiplying $2$,

we can observe that every time for increasing the power of $2$ by $3$, the number of digits increase by $1$.

So to reach $1000th $ power of $2$ we need to add the power by $999$. So, the occurrence of power increasing by $3$ is occurring $333$ times, thus we have $333$ more digits. $2$ is a single digit number. Therefore, we have $334$ digits in total.

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  • $\begingroup$ The question is 'how many digits does $2^{100}$ has', not $2^{1000}$. Don't worry, you don't need to change that much to your answer. $\endgroup$
    – Toby Mak
    Mar 4, 2018 at 5:45

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