Difference between revisions of "2011 AMC 12A Problems/Problem 19"
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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=2^{m+1}-19</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=2^{m+1}-19</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> | + | 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>45</math> and <math>109</math>, that is, <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 == | == Solution 2 == | ||
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== Solution 3 (using the answer choices) == | == Solution 3 (using the answer choices) == | ||
Note that each <math>N</math> is <math>19</math> less than a power of <math>2</math>. So, the answer will be <math>38</math> less than the sum of <math>2</math> powers of <math>2</math>. Adding <math>38</math> to each answer, we get <math>76</math>, <math>128</math>, <math>192</math>, <math>444</math>, and <math>1062</math>. Obviously we can take out <math>76</math> and <math>1062</math>. Also, <math>128</math> will not work because two powers of two will never sum to another power of <math>2</math> (unless they are equal, which is a contradiction to the question). So, we have <math>192</math> and <math>444</math>. Note that <math>444 = 1 + 443 = 2 + 442 = 4 + 440 = 8 + 436 = 16 + 428 = 32 + 412</math>, etc. We quickly see that <math>444</math> will not work, leaving <math>192</math> which corresponds to <math>\boxed{\textbf{C}}</math>. We can also confirm that this works because <math>192 = 128 + 64 = 2^7 + 2^6</math>. | Note that each <math>N</math> is <math>19</math> less than a power of <math>2</math>. So, the answer will be <math>38</math> less than the sum of <math>2</math> powers of <math>2</math>. Adding <math>38</math> to each answer, we get <math>76</math>, <math>128</math>, <math>192</math>, <math>444</math>, and <math>1062</math>. Obviously we can take out <math>76</math> and <math>1062</math>. Also, <math>128</math> will not work because two powers of two will never sum to another power of <math>2</math> (unless they are equal, which is a contradiction to the question). So, we have <math>192</math> and <math>444</math>. Note that <math>444 = 1 + 443 = 2 + 442 = 4 + 440 = 8 + 436 = 16 + 428 = 32 + 412</math>, etc. We quickly see that <math>444</math> will not work, leaving <math>192</math> which corresponds to <math>\boxed{\textbf{C}}</math>. We can also confirm that this works because <math>192 = 128 + 64 = 2^7 + 2^6</math>. | ||
+ | |||
+ | == Solution 4 (removing the log) == | ||
+ | In order to fix the exponent and get rid of the logarithm term, let <math>N = 2^m + k + 1</math>, with <math>0 \leq k < 2^m</math>. Doing so, we see that <math>\lfloor \log_2{N - 1} \rfloor = m</math>, which turns our given relation into | ||
+ | <cmath>2^m = 20 + k,</cmath> | ||
+ | for which the solutions of the form <math>(m, k)</math>, <math>(5, 12)</math> and <math>(6, 44)</math>, follow trivially. Adding up the two values of <math>N</math> gives us <math>32 + 12 + 1 + 64 + 44 + 1 = 154</math>, so the answer is <math>\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}} |
Latest revision as of 10:12, 3 December 2023
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 , that is, 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
Solution 3 (using the answer choices)
Note that each is less than a power of . So, the answer will be less than the sum of powers of . Adding to each answer, we get , , , , and . Obviously we can take out and . Also, will not work because two powers of two will never sum to another power of (unless they are equal, which is a contradiction to the question). So, we have and . Note that , etc. We quickly see that will not work, leaving which corresponds to . We can also confirm that this works because .
Solution 4 (removing the log)
In order to fix the exponent and get rid of the logarithm term, let , with . Doing so, we see that , which turns our given relation into for which the solutions of the form , and , follow trivially. Adding up the two values of gives us , so the answer is .
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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