Difference between revisions of "2011 AMC 12A Problems/Problem 19"
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== Problem == | == Problem == | ||
| + | At a competition with <math>N</math> players, the number of players given elite status is equal to <math>2^{1+\lfloor \log_{2} (N-1) \rfloor}-N</math>. Suppose that <math>19</math> players are given elite status. What is the sum of the two smallest possible values of <math>N</math>? | ||
| + | |||
| + | <math> | ||
| + | \textbf{(A)}\ 38 \qquad | ||
| + | \textbf{(B)}\ 90 \qquad | ||
| + | \textbf{(C)}\ 154 \qquad | ||
| + | \textbf{(D)}\ 406 \qquad | ||
| + | \textbf{(E)}\ 1024 </math> | ||
| + | |||
== Solution == | == Solution == | ||
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=18|num-a=20|ab=A}} | {{AMC12 box|year=2011|num-b=18|num-a=20|ab=A}} | ||
Revision as of 02:36, 10 February 2011
Problem
At a competition with 
players, the number of players given elite status is equal to 
. Suppose that 
players are given elite status. What is the sum of the two smallest possible values of ?
Solution
See also
| 2011 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 18 |
Followed by Problem 20 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
