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 $2$ 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 because $2^{a_7-27}$ 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

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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2022 AMC 12B Problems/Problem 9

Problem

Solution 1

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