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Contributors provided checks that $99^n$ has $2n$ digits and $999^n$ has $3n$ digits for many values of $n$, and this was enough for some people to believe that it is always so, but that is definitely not true.
Soh Yong Sheng, age 12, of Tao Nan School, Singapore argued that the first digits of the powers of 99 get smaller as the powers increase so you would expect that eventually there are less than $2n$ digits.
Although, $99^n$ has $2n$ digits for $n \leq 229$, this is not true for $n \geq 230$. Similarly $999^n$ has $3n$ digits for $n \leq 2301$ but this is not true $n \geq 2302$.
Sergio Moya and Ling Xiang Ning used the following method. To prove the result for $99^n$, note that the smallest whole number with $2n$ digits is $10^{2n-1}$ (for example, the smallest whole number with 6 digits is a hundred thousand, which is $10^5$). So If $99^n$ has $2n$ digits then we must have $99^n \geq 10^{2n-1}$, and this means that we must have
$\begin{eqnarray} \\ 10 \geq \frac{10^{2n}}{99^n} &=& \left( \frac{100}{99} \right)^n.\quad (1) \end{eqnarray}$
If we check this on a calculator for $n=230$ we see that
$$ \left({100\over 99}\right)^{230} = 10.090\cdots $$
so that (1) is false. We can see, then, that $99^{230}$ does NOT have $2n$ digits. One can check that $99^n$ does have $2n$ digits for $1 \leq n \leq 229$.
A similar argument shows that $999^n$ does not have $3n$ digits when $n= 2302$.
Michael Swarbrick-Jones, age 12, Comberton Village College, Cambridge proved the result using logarithms to the base 10. If you know about logarithms, you should be able to see that (1) is the same as
$$\log 10 \geq n \log \left({100\over 99}\right),$$
which gives
$$n \leq {\log 10\over \log (100/99)} = 229.1\cdots .$$
A similar argument shows that if $999^n$ has $3n$ digits, then
$$n \leq {\log 10\over \log (1000/999)} = 2301.4\cdots .$$
Can you show that $9^n$ has $n$ digits when $n\leq 21$ but not otherwise? You may be able to do this directly on your calculator.
Can you show that $87^n$ has $2n$ digits when $n \leq 16$ but not otherwise?
Rachel's Problem
Age 14 to 16
Challenge Level 
Contributors provided checks that $99^n$ has $2n$ digits and $999^n$ has $3n$ digits for many values of $n$, and this was enough for some people to believe that it is always so, but that is definitely not true.
Soh Yong Sheng, age 12, of Tao Nan School, Singapore argued that the first digits of the powers of 99 get smaller as the powers increase so you would expect that eventually there are less than $2n$ digits.
Although, $99^n$ has $2n$ digits for $n \leq 229$, this is not true for $n \geq 230$. Similarly $999^n$ has $3n$ digits for $n \leq 2301$ but this is not true $n \geq 2302$.
Sergio Moya and Ling Xiang Ning used the following method. To prove the result for $99^n$, note that the smallest whole number with $2n$ digits is $10^{2n-1}$ (for example, the smallest whole number with 6 digits is a hundred thousand, which is $10^5$). So If $99^n$ has $2n$ digits then we must have $99^n \geq 10^{2n-1}$, and this means that we must have
$\begin{eqnarray} \\ 10 \geq \frac{10^{2n}}{99^n} &=& \left( \frac{100}{99} \right)^n.\quad (1) \end{eqnarray}$
If we check this on a calculator for $n=230$ we see that
$$ \left({100\over 99}\right)^{230} = 10.090\cdots $$
so that (1) is false. We can see, then, that $99^{230}$ does NOT have $2n$ digits. One can check that $99^n$ does have $2n$ digits for $1 \leq n \leq 229$.
A similar argument shows that $999^n$ does not have $3n$ digits when $n= 2302$.
Michael Swarbrick-Jones, age 12, Comberton Village College, Cambridge proved the result using logarithms to the base 10. If you know about logarithms, you should be able to see that (1) is the same as
$$\log 10 \geq n \log \left({100\over 99}\right),$$
which gives
$$n \leq {\log 10\over \log (100/99)} = 229.1\cdots .$$
A similar argument shows that if $999^n$ has $3n$ digits, then
$$n \leq {\log 10\over \log (1000/999)} = 2301.4\cdots .$$
Can you show that $9^n$ has $n$ digits when $n\leq 21$ but not otherwise? You may be able to do this directly on your calculator.
Can you show that $87^n$ has $2n$ digits when $n \leq 16$ but not otherwise?
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