How many digits is $99^{99}$ without a calculator?

Question

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: $\left( 100-1\right) ^{99}$

Answer 1

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.

Answer 2

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.

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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 $\varepsilon < \frac{1}{99}$.

Answer 3

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!

Answer 4

Note that $$\left(1-\frac1{100}\right)^{100} $$ is a fairly good approximation for $e$, hence $$99^{99}=100^{99}\cdot\left(1-\frac1{100}\right)^{99}\approx 10^{198}\cdot \frac 1{0.99e} \approx 3\cdot10^{197}.$$

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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 $$10^{197}<\frac14\cdot10^{198}<99^{99}<10^{198} $$

Answer 5

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 $n\ge1$. We'll take for granted also the (generous!) inequality $e\lt10$. Thus

$$\left(100\over99\right)^{99}=\left(1+{1\over99}\right)^{99}\lt e\lt10$$

so $100^{99}\lt10\cdot99^{99}$. Since $100^{99}=10^{198}=10\cdot10^{197}$, the inequality $10^{197}\lt99^{99}$ now follows.