Difference between revisions of "2011 AMC 12A Problems/Problem 19"
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\textbf{(E)}\ 1024 </math> | \textbf{(E)}\ 1024 </math> | ||
− | == Solution == | + | == Solution 1== |
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} \left(\frac{N+19}{2}\right)</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} \left(\frac{N+19}{2}\right)</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>. | 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 == | ||
+ | We examine the value that <math> 2^{1+\lfloor\log_{2}(N-1)\rfloor}</math> takes over various intervals. The <math>\lfloor\log_{2}(N-1)\rfloor</math> 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 <math> 2^{1+\lfloor\log_{2}(N-1)\rfloor} - N</math> is the difference between the next power of 2 above <math> 2^{\lfloor\log_{2}(N-1)\rfloor}</math> and <math>N</math>. We are looking for <math>N</math> such that this difference is 19. The first two <math>N</math> that satisfy this are <math>45 = 64-19</math> and <math>109=128-19</math> for a final answer of <math>45 + 109 = 154 \rightarrow \boxed{\textbf{C}}</math> | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | == 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 == | == 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}} |
Latest revision as of 11:12, 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 |
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All AMC 12 Problems and Solutions |
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