Number of digits in 99 to the 99

Question

How many decimal digits does $99^{99}$ have?

Short answer

$198$ . Write $99^{99}=10^{198}\left(1-\tfrac1{100}\right)^{99}$ , and the bracket is about $e^{-1}\approx0.37$ . So $99^{99}\approx 3.7\times10^{197}$ , which has $198$ digits.

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Lessons behind this question

Counting and Combinatorics

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