2007 AIME II Problem 12

Question

The increasing geometric sequence

x_0,

x_1,

x_2,

\ldots

consists entirely of integral powers of

3.

Given that

\sum_{n=0}^{7} \log_3(x_n) = 308

and

56 \le \log_3\left(\sum_{n=0}^{7} x_n\right) \le 57,

find

\log_3(x_{14}).

Answer

Write

x_n = 3^{a+bn}

with integers

a

and

b \ge 1;

the log-sum condition gives

2a + 7b = 77

The sum

x_0 + \cdots + x_7

lies strictly between

x_7

and

3 x_7,

so its base-

3

log pins down

a + 7b

Every term is a power of

3

and the ratio is a quotient of powers of

3,

so

x_n = 3^{a + bn}

for integers

a

and

b,

with

b \ge 1

since the sequence increases. The first condition gives

\sum_{n=0}^{7} (a + bn) = 8a + 28b = 308,

i.e.

2a + 7b = 77.

For the second condition,

x_7 = 3^{a+7b}

is the largest term, and

\begin{aligned} x_7 &\lt \sum_{n=0}^{7} x_n \\ &\lt x_7\left(1 + \tfrac{1}{3} + \tfrac{1}{9} + \cdots\right) \\ &= \tfrac{3}{2}\,x_7 \lt 3x_7, \end{aligned}

so

\log_3\left(\sum x_n\right)

lies strictly between

a + 7b

and

a + 7b + 1.

The given bounds then force

a + 7b = 56.

Subtracting from

2a + 7b = 77

yields

a = 21,

then

b = 5.

Therefore

\log_3(x_{14}) = a + 14b = 21 + 70

= 91.

Problem 12 in other years