2011 AMC 12A Problem 19

At a competition with

N

players, the number of players given elite status is equal to

2^{1 + \lfloor \log_2 (N - 1) \rfloor} - N.

Suppose that

19

players are given elite status. What is the sum of the two smallest possible values of

N?

Note:

\lfloor x \rfloor

is the greatest integer less than or equal to

x.

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Solution

Let

m = \lfloor \log_2 (N - 1) \rfloor,

so the elite count is

2^{m+1} - N = 19,

giving

N = 2^{m+1} - 19.

Consistency requires

2^m \le N - 1 = 2^{m+1} - 20,

i.e.

2^m \ge 20,

m \ge 5.

The two smallest choices are

m = 5

giving

N = 64 - 19 = 45,

and

m = 6

giving

N = 128 - 19 = 109.

Their sum is

154.

Thus, the correct answer is C.

Tagged: floor and ceiling functions · power of 2 · casework