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I know the answer is $198$.

I realise that if $\log _{10}\left( x\right) =y$, the number $x$ has $\lfloor y\rfloor -1$ digits

So I tried $\log ^{\ }_{10}\left( 99^{99}\right) $ = $\log _{10}\left( 100\left( 1-\dfrac {1}{100}\right) \right) $ = $198 + 99\log_{10}\left( 1-\dfrac {1}{100}\right) $ , then I do not know how to proceed. I guess using this method amounts to finding a good approximation to $\log _{10}\left( 99\right)$

I would also be interested to know how one can solve this with the binomial theorem: $\left( 100-1\right) ^{99}$

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    $\begingroup$ $\log_{10}(99)\approx\log_{10}(100)=2$. $\log_{10}(99)\approx 1.9956$, the difference is small. $\endgroup$ – poyea Jun 4 '18 at 14:42
  • $\begingroup$ Indeed. I would start by thinking about $100^{100}$, which is overshooting it slightly. $\endgroup$ – Robert Soupe Jun 4 '18 at 15:23
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In order to show that $99^{99}$ has $198$ digits, we need to show that

$$10^{197}\le99^{99}\lt10^{198}$$

The second inequality is obvious, since $99^{99}\lt100^{99}=10^{198}$. So it remains to prove the first inequality.

Toward this end, recall that

$$\left(1+{1\over n}\right)^n\lt e$$

for any $n\ge1$. It follows that

$$\left(100\over99\right)^{99}=\left(1+{1\over99}\right)^{99}\lt e\lt3$$

Consequently

$$10\cdot10^{197}=10^{198}\lt3\cdot99^{99}$$

and thus

$$10^{197}\lt\left(10\over3\right)10^{197}\lt99^{99}$$

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Using the Binomial Theorem would be very tedious. Although this doesn't provide as good an estimate as $198$ digits, Bernoulli's Inequality is very quick. $$99^{99}=100^{99}\left(1-\frac1{100}\right)^{99}\ge100^{99}\left(1-\frac{99}{100}\right)=100^{97}=10^{194}$$ so we are certain that $99^{99}$ has at least $194$ digits.

We can also bound above using $$99^{99}<100^{99}=10^{198}$$ so we know that $99^{99}$ has at most $198$ digits.

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The number $x$ has $\lfloor \log_{10} x \rfloor + 1$ digits. Hence $x = 99^{99}$ has $\lfloor 99 \log_{10}(99) \rfloor + 1$ digits. The number $\log_{10}(99)$ is just less than $\log_{10}(100) = 2$, so $\lfloor 99 \log_{10}(99) \rfloor$ is one less than $99 \cdot 2 = 198$, which is $197$. Adding 1 gives the number of digits of $x$: 198.


Edit: Yes, "just less" needs to be quantified, or else things could go wrong, and this sends us down the rabbit hole of good approximations for $\log_{10}(99)$.

Realistically, in a algebra/precalculus setting as the question's tag indicates, this might be justified heuristically by looking at how flat the graph of $\log_{10}(x)$ is near $x = 99$. Wave your hands and say, "good enough."

If we're going down the rabbit hole, other answers give methods; while they are elementary, I'd argue that they're not obvious, especially to a typical algebra/precalculus student. Allowing calculus, there's a simpler/more obvious (IMHO) method. Let "just less" be $\varepsilon$. By a corollary to the mean value theorem, $$ \varepsilon = \log_{10}(100) - \log_{10}(99) \leq (100-99) \max_{x \in [99,100]} \frac{1}{x \ln(10)} = \frac{1}{99 \ln(10)}. $$ This gives the desired quantitative estimate: $\lfloor 99(2 - \varepsilon) \rfloor = \lfloor 198 - 99 \varepsilon \rfloor = 197$, where the last equality follows from the above inequality showing that $\varepsilon < \frac{1}{99}$.

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    $\begingroup$ The "just less" should be justified. It might be "a tad bit more than just a little less" $\endgroup$ – Hagen von Eitzen Jun 4 '18 at 14:56
  • $\begingroup$ I disagree with @HagenvonEitzen; obviously $99<100$ so $\log99<\log100$—and since you’re dealing with the floor function, how much less it is is irrelevant. [+1] $\endgroup$ – gen-z ready to perish Jun 4 '18 at 15:05
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    $\begingroup$ @ChaseRyanTaylor Note that in general, $\lfloor n x\rfloor \ne n\lfloor x\rfloor$. How much less is highly relevant. How much relevant? Remove just $0.3\%$ from $\log_{10}(99)$ and you're wrong. $\endgroup$ – Jean-Claude Arbaut Jun 4 '18 at 15:28
  • $\begingroup$ @Jean-ClaudeArbaut Ohhh, now I see! Good point $\ddot\smile$ In my mind I was thinking backwards; I was focusing on how close $\log99$ is to $\log100$, not how far away it is. $\endgroup$ – gen-z ready to perish Jun 4 '18 at 17:21
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Note that $$\left(1-\frac1{100}\right)^{100} $$ is a fairly good approximation for $e$, hence $$99^{99}=100^{99}\cdot\left(1-\frac1{100}\right)^{99}\approx 10^{198}\cdot \frac 1{0.99e} \approx 3\cdot10^{197}.$$


We can make thge "fairly good" a little more precise (but weaker) with Bernoulli's inequality: $$ 1>\left(1-\frac1{100}\right)^{99}>\left(1-\frac1{100}\right)^{100}=\left(\left(1-\frac1{100}\right)^{50}\right)^2\ge \left(\frac12\right)^2=\frac14$$ so that $$10^{197}<\frac14\cdot10^{198}<99^{99}<10^{198} $$

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Besides $$\log ^{\ }_{10}\left( 99^{99}\right)=99\log ^{\ }_{10}\left( 99\right)\approx 99\log_{10}(100)=99\times2$$, you can also try $$\left( 100-1\right) ^{99}=100^{99}-99\,\cdot\,100^{98}+\tbinom{99}{2}100^{97}+\cdots=100^{99}+\tbinom{99}{2}100^{97}+\cdots-\cdots$$ By observation (since you don't have a calculator), $100^{99}=10^{198}$ is sitting in front, while other terms come later will take care of each other, or you can say $10^{198}$ is dominant among pieces. Therefore, the digits is $198$. It also agrees with the estimation!

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