1993 AMC 12 Problem 19

Attempt Problem 19 of the 1993 AMC 12 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 1993 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

19.

How many ordered pairs

(m,n)

of positive integers are solutions to

\frac4m+\frac2n=1?

1

2

3

4

more than

4

Clear denominators and complete a product involving

m-4

and

n-2

The equation is equivalent to

(m-4)(n-2)=8

Clearing denominators and rearranging gives

\begin{aligned} mn-4n-2m=0 &\quad\Longleftrightarrow\quad\\ (m-4)(n-2)=8. \end{aligned}

Both factors must be positive. The four ordered positive factor pairs of

8

give

\begin{aligned} (m,n)={}&(5,10),(6,6),\\ &(8,4),(12,3). \end{aligned}

Thus the correct answer is D.

Problem 19 in other years