Difference between revisions of "2011 AMC 12A Problems/Problem 19"
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== Solution == | == Solution == | ||
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+ | <math> 2^{1+\lfloor\log_{2}(N-1)\rfloor}-N = 19 </math> | ||
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+ | <math> 2^{1+\lfloor\log_{2}(N-1)\rfloor} = N+19 </math> | ||
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+ | <math> 1+\lfloor\log_{2}(N-1)\rfloor = \log_{2}(N+19) </math> | ||
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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>. | ||
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+ | <math> N=2^{m+1}-19 </math> | ||
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+ | <math> m \le \lfloor\log_{2}(N-1) < m+1 </math> | ||
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+ | <math> 2^{m}+1 \le N < 2^{m+1}+1 </math> | ||
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+ | <math> 2^{m}+1 \le 2^{m+1}-19 < 2^{m+1}+1 </math> | ||
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+ | <math> 2^{m}+20 \le 2^{m+1} < 2^{m+1}+20 </math> | ||
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+ | The two smallest possible value of <math> m </math> where <math> m </math> is a positive integers are <math> 5 </math> and <math> 6 </math> respectively. | ||
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+ | Sum of the two smallest possible value of <math> m = 2^{5+1}-19+2^{6+1}-19=154 </math> | ||
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== 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 03:03, 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
Since is a positive integer, must be in the form of for some positive integer .
The two smallest possible value of where is a positive integers are and respectively.
Sum of the two smallest possible value of
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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All AMC 12 Problems and Solutions |