2009 AIME II Problem 11

Question

For certain pairs

(m, n)

of positive integers with

m \ge n

there are exactly

50

distinct positive integers

k

such that

|\log m - \log k| \lt \log n.

Find the sum of all possible values of the product

mn.

Answer

Answer: 125

Concepts:logarithm counting integers in a range casework

Difficulty rating: 2990

Small hint:

The inequality says exactly that

\frac{m}{n} \lt k \lt mn

Big hint:

Write

m = nq + r

with

0 \le r \lt n;

the count of such

k

is

mn - q - 1 = 50,

so

q(n^2 - 1) + nr = 51

leaves only a few cases

Solution:

The inequality

|\log m - \log k| \lt \log n

is equivalent to

\frac{m}{n} \lt k \lt mn.

Write

m = nq + r

with

0 \le r \lt n;

since

m \ge n \ge 2

(for

n = 1

no

k

works),

q \ge 1.

The integers

k

in the interval are

q + 1,

q + 2,

\ldots,

mn - 1,

so there are

mn - q - 1 = 50

of them, that is,

mn - q = 51,

or

q(n^2 - 1) + nr = 51.

For

n \ge 8

the left side is at least

63,

so

2 \le n \le 7.

Checking each case, only

n = 2,

r = 0,

q = 17

(so

m = 34

) and

n = 3,

r = 1,

q = 6

(so

m = 19

) work. These give

mn = 68

and

mn = 57;

indeed

17 \lt k \lt 68

and

\frac{19}{3} \lt k \lt 57

each contain exactly

50

integers.

The sum of all possible values of

mn

is

68 + 57 = 125.