Difference between revisions of "2011 AMC 12A Problems/Problem 19"

Latest revision as of 10:12, 3 December 2023

Problem

At a competition with $N$ players, the number of players given elite status is equal to $2^{1+\lfloor \log_{2} (N-1) \rfloor}-N$ . Suppose that $19$ players are given elite status. What is the sum of the two smallest possible values of $N$ ?

$\textbf{(A)}\ 38 \qquad \textbf{(B)}\ 90 \qquad \textbf{(C)}\ 154 \qquad \textbf{(D)}\ 406 \qquad \textbf{(E)}\ 1024$

Solution 1

We start with $2^{1+\lfloor\log_{2}(N-1)\rfloor}-N = 19$ . After rearranging, we get $\lfloor\log_{2}(N-1)\rfloor = \log_{2} \left(\frac{N+19}{2}\right)$ .

Since $\lfloor\log_{2}(N-1)\rfloor$ is a positive integer, $\frac{N+19}{2}$ must be in the form of $2^{m}$ for some positive integer $m$ . From this fact, we get $N=2^{m+1}-19$ .

If we now check integer values of N that satisfy this condition, starting from $N=19$ , we quickly see that the first values that work for $N$ are $45$ and $109$ , that is, $2^6-19$ and $2^7 -19$ , giving values of $5$ and $6$ for $m$ , respectively. Adding up these two values for $N$ , we get $45 + 109 = 154 \rightarrow \boxed{\textbf{C}}$

Solution 2

We examine the value that $2^{1+\lfloor\log_{2}(N-1)\rfloor}$ takes over various intervals. The $\lfloor\log_{2}(N-1)\rfloor$ 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 $2^{1+\lfloor\log_{2}(N-1)\rfloor} - N$ is the difference between the next power of 2 above $2^{\lfloor\log_{2}(N-1)\rfloor}$ and $N$ . We are looking for $N$ such that this difference is 19. The first two $N$ that satisfy this are $45 = 64-19$ and $109=128-19$ for a final answer of $45 + 109 = 154 \rightarrow \boxed{\textbf{C}}$

Solution 3 (using the answer choices)

Note that each $N$ is $19$ less than a power of $2$ . So, the answer will be $38$ less than the sum of $2$ powers of $2$ . Adding $38$ to each answer, we get $76$ , $128$ , $192$ , $444$ , and $1062$ . Obviously we can take out $76$ and $1062$ . Also, $128$ will not work because two powers of two will never sum to another power of $2$ (unless they are equal, which is a contradiction to the question). So, we have $192$ and $444$ . Note that $444 = 1 + 443 = 2 + 442 = 4 + 440 = 8 + 436 = 16 + 428 = 32 + 412$ , etc. We quickly see that $444$ will not work, leaving $192$ which corresponds to $\boxed{\textbf{C}}$ . We can also confirm that this works because $192 = 128 + 64 = 2^7 + 2^6$ .

Solution 4 (removing the log)

In order to fix the exponent and get rid of the logarithm term, let $N = 2^m + k + 1$ , with $0 \leq k < 2^m$ . Doing so, we see that $\lfloor \log_2{N - 1} \rfloor = m$ , which turns our given relation into $2^m = 20 + k,$ for which the solutions of the form $(m, k)$ , $(5, 12)$ and $(6, 44)$ , follow trivially. Adding up the two values of $N$ gives us $32 + 12 + 1 + 64 + 44 + 1 = 154$ , so the answer is $\boxed{\textbf{C}}$ .

@@ Line 35: / Line 35: @@
 == 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>.
+== 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 ==
 {{AMC12 box|year=2011|num-b=18|num-a=20|ab=A}}
 {{MAA Notice}}

2011 AMC 12A (Problems • Answer Key • Resources)
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