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1996 AIME Problems/Problem 2

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Problem

For each real number x, let \lfloor x \rfloor denote the greatest integer that does not exceed x. For how many positive integers n is it true that n<1000 and that \lfloor \log_{2} n \rfloor is a positive even integer?

Solution

For integers k, we want \lfloor \log_2 n\rfloor = 2k, or 2k \le \log_2 n < 2k+1 \Longrightarrow 2^{2k} \le n < 2^{2k+1}. Thus, n must satisfy these inequalities (since n < 1000):

4\leq n <8
16\leq n<32
64\leq n<128

256\leq n<512

There are 4 for the first inequality, 16 for the second, 64 for the third, and 256 for the fourth, so the answer is 4+16+64+256=\boxed{340}.

See also

1996 AIME (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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