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How many digits are there in the number $200^{2010}$? I have tried to re-write it as $(2\cdot 100)^{2010} = (2\cdot 10^2)^{2010} = 2^{2010} \cdot 10^{4020} = 1024^{201} \cdot 10^{4020} = 1.024^{201} \cdot 10^{4623}$. But how do I write $1.024^{201}$ as a power of base 10? I would like to solve the problem without logarithms or a calculator.

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  • $\begingroup$ @AndréNicolas - a bit to small. $\endgroup$ – Mark Bennet Sep 5 '13 at 15:02
  • $\begingroup$ Of course. Idea right, detail not so good! $\endgroup$ – André Nicolas Sep 5 '13 at 15:04
  • $\begingroup$ For estimate of $(1.024)^{201}$, use $(1.024)^n\approx 1+0.024n$ if $n$ is not large. So picking $n$ about $40$ we get $2$, then take fifth power. Underestimate, but with a bit of fooling around we can pin things down. $\endgroup$ – André Nicolas Sep 5 '13 at 15:12
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$\log 200^{2010}=2010\times\log 200=2010\times(2+0.3010)=4625.01$

$\therefore$ $4625\leq \log 200^{2010} <4626$

$\therefore$ $10^{4625}\leq 200^{2010}<10^{4626}$

$\therefore$ $4626$ digits

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Think about some very simple cases $10$ has $2$ digits as does $99$, but $100$ has $3$ digits.$1\le \log_{10} 99 \lt 2$. A few explorations should convince you that the number of decimal digits of a positive integer $N$ is $\lfloor \log_{10} N\rfloor +1$.

You can prove this by writing $N=a\times 10^k$ where $1\le a\lt 10$ and $k$ is an integer and taking logs to base 10.

Now $\log_{10} (200^{2010})=2010 \log_{10}(200)=2010 \cdot(2+\log_{10}2)$ so you just need to have a sufficiently accurate estimate of $\log_{10}2$.

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$1024^{201} \approx (10^3)^{201}=10^{603}$ is a rough estimate for that term.

We can do slightly better by writing $1024=10^3 + 24$, so

$1024^{201} = (10^3 + 24)^{201} = 10^{603} + 201\cdot24\cdot10^{600} + \tbinom {201}2\cdot24^2\cdot10^{597} + \cdots$

After just a few terms, they should be small enough so as not to add digits, and you can probably work that out without too much trouble.

This is kind of clunky, but it does completely avoid logarithms, if that's important.

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You cannot really solve it without logarithms since this is exactly what logarithm is – it measures how many digits you need to write a number in a certain base. The number of decimal digits is $⌊\log_{10}(n)⌋ + 1$. $\log(200^{2010}) = 2010 * \log(200) = 2010 * \log(2 * 100) = 2010 * (\log(2) + 2) = 4625.07…$. So the exact number of digits is 4626. If you don't want to use calculator, you can estimate $\log_{10}(2)$ by another method.

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  • $\begingroup$ Which means the approximation $1000 \approx 1024$ throws you off by two digits. I'd say that's a win. $\endgroup$ – Arthur Sep 5 '13 at 15:03
  • $\begingroup$ @Arthur: Yes, nice. $\endgroup$ – user87690 Sep 5 '13 at 15:08

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