2007 AMC 10B Problem 25

Question

How many pairs of positive integers

(a,b)

are there such that

a

and

b

have no common factors greater than

1

and

\frac{a}{b}+\frac{14b}{9a}

is an integer?

4

6

9

12

infinitely many

Combine into the single fraction

\dfrac{9a^2+14b^2}{9ab}

and use

\gcd(a,b)=1

Deduce that

14

is divisible by

a

and

9

is divisible by

b,

then test the finitely many pairs

Combining, the expression is

\dfrac{9a^2+14b^2}{9ab}.

For this to be an integer,

a

must divide

9a^2+14b^2,

hence

14b^2

is divisible by

a.

Since

\gcd(a,b)=1,

we get that

14

is divisible by

a.

Similarly,

9a^2

is divisible by

b,

so

9

is divisible by

b.

So

a\in\{1,2,7,14\}

and

b\in\{1,3,9\}.

Checking these, only

b=3

makes the expression an integer for each allowed

a.

The valid pairs are

(1,3),

(2,3),

(7,3),

and

(14,3),

for a total of

4.

Thus, the correct answer is A.

Problem 25 in other years