2011 AMC 12A Problems/Problem 19
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 |
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