If \log_{10} 2 = 0.301 and \log_{10} 3 = 0.477, then what is the number of digits in the expansion of 60^{60}?

Explanation

Let N = 60^{60}. Taking log: \log_{10} N = 60 \log_{10} 60 = 60(\log_{10} 10 + \log_{10} 2 + \log_{10} 3) = 60(1 + 0.301 + 0.477) = 60(1.778) = 106.68. The number of digits is the integer part + 1, which is 106 + 1 = 107.

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