2009/10 AHSMC Part 1 Problem 8

Problem 8 of 16

If $2^{2009}$ has $m$ digits and $5^{2009}$ has $n$ digits in their base- $10$ representations, then the value of $m + n$ is

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Solution

Since $10^{m-1} < 2^{2009} < 10^m$ and $10^{n-1} < 5^{2009} < 10^n$ , we have $10^{m+n-2} < 2^{2009}5^{2009} < 10^{m+n}$ . It follows that $10^{m+n-1} = 2^{2009}5^{2009} = 10^{2009}$ . Hence $m + n - 1 = 2009$ . The answer is (D).

University of Alberta

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