2022 AMC 12B Problem 9

Question

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

Divide both sides by

2^{27}

to get

2^{a_7 - 27} = a_7

After finding

a_7,

write

a_2 = a_7 - 5d

and make the common difference

d

as large as possible

Dividing by

2^{27},

we need

2^{a_7 - 27} = a_7.

Thus

a_7=2^j

and

2^j-j=27,

where

j=a_7-27\ge0.

The left side is strictly increasing for

j\ge1,

and

j=5

works, so uniquely

a_7=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.

Problem 9 in other years