Difference between revisions of "2011 AMC 12A Problems/Problem 19"
Latex turtle (talk | contribs) (Added alternate solution) |
Drunkenninja (talk | contribs) m (→Solution 2) |
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− | From this, we see that <math> 2^{1+\lfloor\log_{2}(N-1)\rfloor}</math> is the difference between <math>N</math> and | + | From this, we see that <math> 2^{1+\lfloor\log_{2}(N-1)\rfloor} - N</math> is the difference between the next power of 2 above <math> 2^{\lfloor\log_{2}(N-1)\rfloor}</math> and <math>N</math>. We are looking for <math>N</math> such that this difference is 19. The first two <math>N</math> that satisfy this are <math>45 = 64-19</math> and <math>109=128-19</math> for a final answer of <math>45 + 109 = 154 \rightarrow \boxed{\textbf{C}}</math> |
== 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:22, 8 January 2018
Contents
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 1
We start with . After rearranging, we get .
Since is a positive integer, must be in the form of for some positive integer . From this fact, we get .
If we now check integer values of N that satisfy this condition, starting from , we quickly see that the first values that work for are and , giving values of and for , respectively. Adding up these two values for , we get
Solution 2
We examine the value that takes over various intervals. The means it changes on each multiple of 2, like so:
2 --> 1
3 - 4 --> 2
5 - 8 --> 3
9 - 16 --> 4
From this, we see that is the difference between the next power of 2 above and . We are looking for such that this difference is 19. The first two that satisfy this are and for a final answer of
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 |
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