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How many digits are there in $2020 ^{2020} $ ?

In solution, I first factorized the given number to be $202^{2020}\times10^{2020}$

This made it sufficient to calculate the total digit number of $202^{2020}$ and then add $2020$ digits (for the zeroes in $10^{2020}$ ) to find the answer

Now I found out all powers of $202$ up to $7th$ power, by hand-multiplication. What I figured out is:

For every $202^{1+3n}$ no. of digits in the answer is $ 3 + 7n $

This way the answer of this question should lead to $6734$ digits where, $202^{2020}$ has $4714$ digits, and $2020$ more digits for $10^{2020}$

My question is, whether the formula I mentioned in bold letter, is always applicable up to any natural number value of $n$ ?

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    Hint: the number of digits in $N$ is $\lceil \log_{10} N\rceil$ (where $\lceil x\rceil$ denotes the least integer which is not less than $x$). – 2017-02-07
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    To your numerical question: No. with $n=4$ we have that $202^{13}$ has $30$ digits but $3+7n=31$ and the gap continues to widen for larger $n$. – 2017-02-07
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    @lulu, can we use a method to solve this problem without calculator? – 2017-02-07
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    An exact answer? Probably, but nothing leaps to mind. To do it roughly, I'd approximate with $2^{2020}\times 10^{6060}$ and since $2^{10}\approx 10^3$ we'd then get around $10^{606}\times 10^{6060}=10^{6666}$ which would mean $6665$. That's not terrible! the correct answer (barring error) is $6677$. – 2017-02-07
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    @lulu To deal correctly with powers of $10$ e.g. $10$ has two digits you need the floor of the log plus one, rather than the ceiling. The functions are the same except at powers of $10$, so it makes no difference to he answer to the question. – 2017-02-07
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    @MarkBennet Thanks for the correction! You are, of course, absolutely correct. – 2017-02-07

2 Answers 2

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Let me use $\log x$ for the logarithm in base $10$ of $x$ and $\ln x$ for the natural logarithm.

You can do the computation by hand if you know $\log2=0.301030$ (which I had to memorize in school), $\ln10=2.302586$ (which I was supposed to memorize but never did) and the approximation $\ln(1+x)\approx x-x^2/2$ for $x$ close to $0$. $$ \log202=\log200+\log\Bigl(1+\frac1{100}\Bigr)\approx2+\log2+\frac{1}{\ln10}\Bigl(\frac{1}{100}-\frac12\frac{1}{100^2}\Bigr). $$

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Your finding

$$202^{1+3n}\to3+7n\text{ digits}$$

is only approximate.

Actually, using logarithms,

$$\log_{10}202^{1+3n}=(1+3n)\log_{10}202=2.3053513694\cdots+6.9160541083\cdots n.$$

Taking the ceiling, the two formulas give

$$0 \to 3 , 3 \\ 1 \to 10 , 10 \\ 2 \to 17 , 17 \\ 3 \to 24 , 24 \\ 4 \to 31 , 30 \\ 5 \to 38 , 37 \\ 6 \to 45 , 44 \\ 7 \to 52 , 51 \\ 8 \to 59 , 58 \\ 9 \to 66 , 65 \\\cdots$$

The correct answer is

$$\lceil2020\log_{10}2020\rceil=6677.$$