2022 AMC 12B Problems/Problem 9

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 $2$ larger than $27$ . The first power of 2 larger than $27$ , namely $32$ , does indeed work. However, if $a_7>32$ , $2^{a_7-27}$ would grow at an exponential rate whereas $a_7$ would grow at a linear rate, so the graphs would 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 $d = 4$ , $a_2 = \fbox{12 (B)}$ , and we're done!

~Bxiao31415

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

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2022 AMC 12B Problems/Problem 9

Problem

Solution 1

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