3
$\begingroup$

Is is possible to find the number of digits of $2013^{2013}$ without a calculator?

$\endgroup$
  • 3
    $\begingroup$ Can't really get more than $3$ digit accuracy out of a slide rule, so will need to go to log tables. $\endgroup$ – André Nicolas Aug 4 '13 at 15:34
3
$\begingroup$

HINT:

Number of digits of $N$ in base $b$ =$1+\lfloor\log_bN\rfloor$ [Proof]

Without calculator $10^3<2013<10^4\implies 3<\log_{10}2013<4$

or as $\sqrt{10^7}>3000>2013, 10^7>2013^2\implies \log_{10}2013<\frac72=3.5$

$\endgroup$
  • 1
    $\begingroup$ Please note that $\lfloor 2013 x \rfloor$ is not necessarily equal to $2013 \lfloor x \rfloor$. You need stronger bounds. $\endgroup$ – Najib Idrissi Aug 4 '13 at 15:38
  • $\begingroup$ @nik, I think the comment is directed to Daniel $\endgroup$ – lab bhattacharjee Aug 4 '13 at 15:39
  • 2
    $\begingroup$ It is not. The bounds you prove are $3 < \log_{10} 2013 < 3.5$. This is not sufficient to get the value of $\lfloor 2013 \log_{10} 2013 \rfloor$. $\endgroup$ – Najib Idrissi Aug 4 '13 at 15:40
  • $\begingroup$ @nik, I don't think without calculator or rigorous calculation, we can achieve better accuracy $\endgroup$ – lab bhattacharjee Aug 4 '13 at 15:42
  • $\begingroup$ Yes, I agree. It's still worth mentioning, I think. $\endgroup$ – Najib Idrissi Aug 4 '13 at 15:43
5
$\begingroup$

$$1+\lfloor\log_{10}(2013^{2013})\rfloor = 1 + \lfloor 2013\log_{10}(2013)\rfloor = 6651$$

A good estimate without using a calculator is given by $$\log_{10}(2013) = \log_{10}(1000\cdot2\cdot1.0065) = \log_{10}(1000) + \log_{10}(2)+\log_{10}(1.0065)$$ which is approximatively $$\log_{10}(2013)\approx 3+\frac{1}{3}+0 \approx 3.35$$ where the exact result is $\log_{10}(2013) = 3.30384\ldots$

$\endgroup$
  • 2
    $\begingroup$ how have you done this "without a calculator"? $\endgroup$ – lab bhattacharjee Aug 4 '13 at 15:43
  • 1
    $\begingroup$ This is not too difficult. The only constant that you need to remember is $\ln 10\approx 2.303$. Then starting from $2^{10}=1024$ you get ($\log$ in base ten) $$10\log 2=3+\log(1.024).$$ Using the differential we get $$\log(1+t)\approx\frac{t}{2.303}$$ for $t\approx0$. This gives $$10\log2=3+\frac{0.024}{2.303}\approx3.0103$$ with pen and paper division of $2400/2303$. The same formula gives $$\log(1.0065)\approx\frac{0.0065}{2.303}\approx0.00282.$$ All this gives $\log2013\approx 3.30385$ (one of us made a mistake in the fifth decimal). Multiplying that is another pen and paper job. $\endgroup$ – Jyrki Lahtonen Aug 4 '13 at 16:27
  • 1
    $\begingroup$ Ok, I did remeber $\log 2$ as well. That formula above rounds up to $0.30104$. But all the above can be done with pen and paper. Feynman would have done this with mental arithmetic alone. $\endgroup$ – Jyrki Lahtonen Aug 4 '13 at 16:33
  • 1
    $\begingroup$ Feynman and exponential. Euler was also famous for his abilities with mental arithmetic. He was blind towards the end, so he had no choice :-) $\endgroup$ – Jyrki Lahtonen Aug 4 '13 at 16:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.