Question

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 $x.$

$38$

$90$

$154$

$406$

$1024$

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. $2^m \ge 20,$ so $m \ge 5.$

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.

2011 AMC 12A Problem 19

Solution:

Problem 19 in Other Years