How many Digits are there in $2020^{2020}$

Question

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 $n$ ?

Answer 1

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 $0$. $$ \log202=\log200+\log\Bigl(1+\frac1{100}\Bigr)\approx2+\log2+\frac{1}{\ln10}\Bigl(\frac{1}{100}-\frac12\frac{1}{100^2}\Bigr). $$

Answer 2

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

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

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The correct answer is

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