Difference between revisions of "2011 AMC 12A Problems/Problem 19"
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− | We start with <math> 2^{1+\lfloor\log_{2}(N-1)\rfloor}-N = 19</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} (\frac{N+19}{2})</math>. |
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− | After rearranging, we get | ||
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− | <math>\lfloor\log_{2}(N-1)\rfloor = \log_{2} (\frac{N+19}{2})</math>. | ||
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Since <math> \lfloor\log_{2}(N-1)\rfloor </math> is a positive integer, <math> \frac{N+19}{2}</math> must be in the form of <math>2^{m} </math> for some positive integer <math> m </math>. From this fact, we get <math>N+19=2^{m+1}</math>. | Since <math> \lfloor\log_{2}(N-1)\rfloor </math> is a positive integer, <math> \frac{N+19}{2}</math> must be in the form of <math>2^{m} </math> for some positive integer <math> m </math>. From this fact, we get <math>N+19=2^{m+1}</math>. |
Revision as of 21:30, 11 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
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
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 |