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Difference between revisions of "2022 AMC 12B Problems/Problem 9"

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== Problem ==
 
== Problem ==
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The sequence <math>a_0,a_1,a_2,\cdots</math> is a strictly increasing arithmetic sequence of positive integers such that<cmath>2^{a_7}=2^{27} \cdot a_7.</cmath>What is the minimum possible value of <math>a_2</math>?
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The sequence <math>a_0,a_1,a_2,\cdots</math> is a strictly increasing arithmetic sequence of positive integers such that <cmath>2^{a_7}=2^{27} \cdot a_7.</cmath> What is the minimum possible value of <math>a_2</math>?
  
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<math>\textbf{(A)}8 \qquad \textbf{(B)}12 \qquad \textbf{(C)}16 \qquad \textbf{(D)}17 \qquad \textbf{(E)}22</math>
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<math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 16 \qquad \textbf{(D)}\ 17 \qquad \textbf{(E)}\ 22</math>
  
 
== Solution 1 ==
 
== Solution 1 ==
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We can rewrite the given equation as <math>2^{a_7-27}=a_7</math>. Hence <math>a_7</math> must be a power of <math>2</math> larger than <math>27</math>. The first power of 2 larger than <math>27</math>, namely <math>32</math>, does indeed work. However, if <math>a_7>32</math>, <math>2^{a_7-27}</math> would grow at an exponential rate whereas <math>a_7</math> would grow at a linear rate, so the graphs would not intersect again.
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We can rewrite the given equation as <math>2^{a_7-27}=a_7</math>. Hence, <math>a_7</math> must be a power of <math>2</math> and larger than <math>27</math>. The first power of 2 that is larger than <math>27</math>, namely <math>32</math>, does satisfy the equation: <math>2^{32 - 27} = 2^5 = 32</math>. In fact, this is the only solution; <math>2^{a_7-27}</math> is exponential whereas <math>a_7</math> is linear, so their graphs will not intersect again.
  
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Now, let the common difference in the sequence be <math>d</math>. Hence <math>a_0 = 32 - 7d</math> and <math>a_2 = 32 - 5d</math>. To minimize <math>a_2</math>, we maxmimize <math>d</math>. Since the sequence contains only positive integers, <math>32 - 7d > 0</math> and hence <math>d \leq 4</math>. When <math>d = 4</math>, <math>a_2 = \fbox{12 (B)}</math>, and we're done!
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Now, let the common difference in the sequence be <math>d</math>. Hence, <math>a_0 = 32 - 7d</math> and <math>a_2 = 32 - 5d</math>. To minimize <math>a_2</math>, we maxmimize <math>d</math>. Since the sequence contains only positive integers, <math>32 - 7d > 0</math> and hence <math>d \leq 4</math>. When <math>d = 4</math>, <math>a_2 = \boxed{\textbf{(B)}\ 12}</math>.
  
 
~[[User:Bxiao31415|Bxiao31415]]
 
~[[User:Bxiao31415|Bxiao31415]]
 +
 +
 +
==Video Solution (Under 2 min!)==
 +
https://youtu.be/N_v7AXXsRfo
 +
 +
~<i>Education, the Study of Everything</i>
 +
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==Video Solution(1-16)==
 +
https://youtu.be/SCwQ9jUfr0g
 +
 +
~~Hayabusa1
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2022|ab=B|num-b=8|num-a=10}}
 
{{AMC12 box|year=2022|ab=B|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:04, 9 July 2023

Problem

The sequence $a_0,a_1,a_2,\cdots$ is a strictly increasing arithmetic sequence of positive integers such that \[2^{a_7}=2^{27} \cdot a_7.\] What is the minimum possible value of $a_2$?

$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 16 \qquad \textbf{(D)}\ 17 \qquad \textbf{(E)}\ 22$

Solution 1

We can rewrite the given equation as $2^{a_7-27}=a_7$. Hence, $a_7$ must be a power of $2$ and larger than $27$. The first power of 2 that is larger than $27$, namely $32$, does satisfy the equation: $2^{32 - 27} = 2^5 = 32$. In fact, this is the only solution; $2^{a_7-27}$ is exponential whereas $a_7$ is linear, so their graphs will not intersect again.

Now, let the common difference in the sequence be $d$. Hence, $a_0 = 32 - 7d$ and $a_2 = 32 - 5d$. To minimize $a_2$, we maxmimize $d$. Since the sequence contains only positive integers, $32 - 7d > 0$ and hence $d \leq 4$. When $d = 4$, $a_2 = \boxed{\textbf{(B)}\ 12}$.

~Bxiao31415


Video Solution (Under 2 min!)

https://youtu.be/N_v7AXXsRfo

~Education, the Study of Everything

Video Solution(1-16)

https://youtu.be/SCwQ9jUfr0g

~~Hayabusa1

See also

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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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