2005 AIME II Problem 5

Attempt Problem 5 of the 2005 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AIME II solutions, or check the answer key.

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5.

Determine the number of ordered pairs

(a, b)

of integers such that

\log_a b + 6\log_b a = 5,

2 \le a \le 2005,

and

2 \le b \le 2005.

Answer

Answer: 54

Concepts:logarithm quadratic counting integers in a range

Difficulty rating: 2310

Small hint:

Set

x = \log_a b;

since

\log_b a = \frac{1}{x},

the equation becomes

x + \frac{6}{x} = 5

Big hint:

The two cases are

b = a^2

and

b = a^3;

count how many

a \ge 2

keep

b \le 2005

in each

Solution:

Let

x = \log_a b.

Since

\log_b a = \frac{1}{x},

the equation becomes

x + \frac{6}{x} = 5,

i.e.

x^2 - 5x + 6 = 0,

so

x = 2

or

x = 3.

That means

b = a^2

or

b = a^3.

For

b = a^2 \le 2005

we need

2 \le a \le 44

(since

44^2 = 1936

and

45^2 = 2025

), giving

43

pairs. For

b = a^3 \le 2005

we need

2 \le a \le 12

(since

12^3 = 1728

and

13^3 = 2197

), giving

11

pairs.

In total there are

43 + 11 = 54

ordered pairs.