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

Najib Idrissi
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Commander Shepard
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2 Answers2

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

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 $\log_{10}(2013) = 3.30384\ldots$

Daniel Robert-Nicoud
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    how have you done this "without a calculator"? – lab bhattacharjee Aug 04 '13 at 15:43
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    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. – Jyrki Lahtonen Aug 04 '13 at 16:27
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    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. – Jyrki Lahtonen Aug 04 '13 at 16:33
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    Feynman and exponential. Euler was also famous for his abilities with mental arithmetic. He was blind towards the end, so he had no choice :-) – Jyrki Lahtonen Aug 04 '13 at 16:43
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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 $\sqrt{10^7}>3000>2013, 10^7>2013^2\implies \log_{10}2013<\frac72=3.5$

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lab bhattacharjee
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