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How could I solve this problem?

Find the first digit of $2^{4242}$ without using a calculator.

I know how to find the last digit with modular arithmetic, but I can't use that here.

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I simply edited the typo in the title. I hope you don't mind! – amWhy Nov 15 '12 at 14:36
@amWhy, Thank you – sperners lemma Nov 15 '12 at 14:37
If you have a table of logarithms, you can evaluate $4242 \log(2)$ with sufficient accuracy to bound its fractionnal part between the logarithm of two consecutive numbers between $1$ and $10$ ($\log(d) \le \{4242 \log(2)\} < \log(d+1)$ yields that the first digit of $2^{4242}$ is $d$). – Joel Cohen Nov 15 '12 at 14:56
Giving a reliable mathematical answer either requires access to some information like the value of $\log(2)$ in at least $4$ decimal places, or quite a bit of computation (you could compute that value by hand). Could this be a trick question? The first digit is the one on the left of the expression, and it is a $2$. – Marc van Leeuwen Nov 15 '12 at 15:04
@MarcvanLeeuwen, the answer is 9. I ask because I want to know about first digits of exponentiated numbers. – sperners lemma Nov 15 '12 at 15:05
up vote 11 down vote accepted

This is probably not the answer you are looking for, and wil probably only be appreciated by people of my age ...

I can still remember from school days that $\log_{10} 2 = 0,30102999$ (I always thought it was noteworthy that it is so close to $0.30103$) - people who went to school in the 1950's can probably recall using logs to base 10 for lots of tedious calculations.

You can then do the multiplication by 4242 without a calculator, and get the fractional part ($=x$, say, but you are likely to need a calculator to find out the first digit of $10^x$, unless you have also memorised $\log 2, \log3, \dots, \log 9$ (I can't!)


With a bit more digging in the recesses of my memory, I can just recall that $\log 3$ is something like $0.477$, so $\log 9 = 2 \log 3 = 0.954$, so that should do it ...

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Oops - I was finding the number of digits - I will correct ... – Old John Nov 15 '12 at 15:02
It would be enough to know $\log 3$ accurately as $\log 9 = 2 \log 3$. Unfortunately, the best I know $\log 3$ is a little less than 0.5, not good enough for this purpose. I suppose you could do a bit of a Taylor series starting from $\log \sqrt 10=0.5$ – Ross Millikan Nov 15 '12 at 15:08

If you have memorized that $\log_{10}2\approx 0.30103$ you can multiply by $4242$ and take the fractional part as $0.9692$ which looks like it should be greater than $\log_{10}9$ (and it is, but it is closer than I would have thought, it is $\approx 0.9542$). I don't know how to do it without log tables.

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To get $0.9692$ you needed more digits than present in $0.30103$. But having cheated with a calculator, I can confirm that the first digit is $9$ (it starts $93162601\ldots$). – Marc van Leeuwen Nov 15 '12 at 15:11
@MarcvanLeeuwen: True, but I only need that it is greater than 0.9542. You are right that these aren't enough digits to know that-the fractional part of $4242*0.301025=0.94805\lt \log 9$. But 0.30103 is three digits better than shown. – Ross Millikan Nov 15 '12 at 15:17
Instead of a new one here +2 for two ones here. – Babak S. Nov 15 '12 at 16:20

Presenting an alternate method (no logs, but needs the knowledge that doubling time is ~ $70$/rate)

Starting with

$$2^{10}=1024$$ which is $$1000 \times 1.024$$ or a 2.4% increase.

Then, $$70/2.4 \approx 29$$ implies that $$2^{290}\sim 2 \times 10^k$$ for some k.

Then, $$2^{4242} = {2^{290}}^{14} \times 2^{182}$$

So, it would suffice to calculate $$2^{16} \approx 1.6 \times 10^l$$ and to get $2^{182}$, first note that $1.024^{29} \approx 2$ so, $1.024^{18} \approx 1.5$ (pure hand waving, but sounds logical) So, from that $$2^{182} \approx 6 \times 10^m$$ and thus we can get the fist digit to be close to $1.6 \times 6 > 9$

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Instead of tildes, which don't show up in the $\LaTeX$, you can use \sim to get $\sim$ – Ross Millikan Nov 15 '12 at 18:46
@Ross, my TeX hasn't been working I have posted about that in meta before. So, I can't really see what I am typing. – picakhu Nov 15 '12 at 18:48
Actually, the statement of doubling time being 70/rate comes from $\ln 2 \approx 0.7$ so you are just hiding the logs. It is also not clear that the accuracy is enough to be sure it is $9$ – Ross Millikan Nov 15 '12 at 18:49
@Ross, I agree with you. But that is more "common knowledge" than knowing $\log 2 = 0.69...$ – picakhu Nov 15 '12 at 18:51

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