Difference between revisions of "2011 AMC 12A Problems/Problem 19"
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\textbf{(E)}\ 1024 </math> | \textbf{(E)}\ 1024 </math> | ||
− | == Solution == | + | == Solution 1== |
We start with <math> 2^{1+\lfloor\log_{2}(N-1)\rfloor}-N = 19</math>. After rearranging, we get <math>\lfloor\log_{2}(N-1)\rfloor = \log_{2} \left(\frac{N+19}{2}\right)</math>. | We start with <math> 2^{1+\lfloor\log_{2}(N-1)\rfloor}-N = 19</math>. After rearranging, we get <math>\lfloor\log_{2}(N-1)\rfloor = \log_{2} \left(\frac{N+19}{2}\right)</math>. | ||
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If we now check integer values of N that satisfy this condition, starting from <math>N=19</math>, we quickly see that the first values that work for <math>N</math> are <math>2^6 -19</math> and <math>2^7 -19</math>, giving values of <math>5</math> and <math>6</math> for <math>m</math>, respectively. Adding up these two values for <math>N</math>, we get <math>45 + 109 = 154 \rightarrow \boxed{\textbf{C}}</math> | If we now check integer values of N that satisfy this condition, starting from <math>N=19</math>, we quickly see that the first values that work for <math>N</math> are <math>2^6 -19</math> and <math>2^7 -19</math>, giving values of <math>5</math> and <math>6</math> for <math>m</math>, respectively. Adding up these two values for <math>N</math>, we get <math>45 + 109 = 154 \rightarrow \boxed{\textbf{C}}</math> | ||
+ | |||
+ | == Solution 2== | ||
+ | We examine the value that <math> 2^{1+\lfloor\log_{2}(N-1)\rfloor}</math> takes over various intervals. The <math>\lfloor\log_{2}(N-1)\rfloor</math> 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 <math> 2^{1+\lfloor\log_{2}(N-1)\rfloor}</math> is the difference between <math>N</math> and the next power of 2 above it. 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 19:45, 9 September 2017
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 and the next power of 2 above it. 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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