2022 AMC 12B Problem 9

Below is the professionally curated solution for Problem 9 of the 2022 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 12B solutions, or check the answer key.

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Concepts:arithmetic sequence exponent optimization

Difficulty rating: 1530

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 $a_2?$

$8$

$12$

$16$

$17$

$22$

Solution:

Dividing by $2^{27},$ we need $2^{a_7 - 27} = a_7.$ The only positive integer solution is $a_7 = 32,$ since $2^{32 - 27} = 2^5 = 32.$

With common difference $d \ge 1,$ we have $a_7 = a_0 + 7d = 32$ and $a_2 = a_0 + 2d = 32 - 5d.$ To minimize $a_2$ we maximize $d;$ since $a_0 = 32 - 7d \ge 1,$ the largest choice is $d = 4$ (giving $a_0 = 4$ ).

Then $a_2 = 32 - 20 = 12.$

Thus, the correct answer is B.

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