How Many Digits Are in $19^{9^9}$?

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    2017
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SUMMARY

The problem of the week (POTW) focuses on determining the number of digits in the decimal representation of the integer $19^{9^9}$. To find the number of digits, one can use the formula for the number of digits in a number, which is given by $\lfloor \log_{10}(n) \rfloor + 1$. Applying this to $19^{9^9}$ involves calculating $\log_{10}(19^{9^9}) = 9^9 \cdot \log_{10}(19)$. The final result reveals that $19^{9^9}$ contains 20 digits.

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Here is this week's POTW:

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How many digits are there in (decimal representation of) the integer $19^{9^9}$?

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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No one answered last week's problem.(Sadface)

You can find the suggested solution below:
Note that $19^{9^9}=10^{9^9\log 19}=10^{9^9 \cdot 1.27875} \approx 10^{495,415,345.4}$, therefore the number $19^{9^9}$ requires $495,415,346$ digits.
 

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