Question

44. For how many integers n is \(\frac{n}{20-n}\) the square of an integer?

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

Set n/(20-n) equal to small perfect squares like 1 and 4, and solve for n each time to find how many integers work.

Answer 1

The Correct Option is C

Step 1: Work out where the fraction can possibly be a perfect square.
The expression \(\frac{n}{20-n}\) is defined for every integer \(n\) except \(n = 20\). Since the square of any integer is never negative, we only need to look at values of \(n\) that make this fraction zero or positive.

Step 2: Rule out n greater than 20.
When \(n > 20\), the term \(20 - n\) turns negative while \(n\) stays positive, so the whole fraction \(\frac{n}{20-n}\) is negative. A negative value can never be the square of an integer, so no integer greater than 20 can work.

Step 3: Focus on 0 < n < 20 and test small perfect squares.
In this range both \(n\) and \(20-n\) are positive, so the fraction is a positive number, and we can check which perfect square it lands on. Set \(\frac{n}{20-n} = 1\) first:
\[ n = 20 - n \implies 2n = 20 \implies n = 10 \]
Check it: \(\frac{10}{20-10} = \frac{10}{10} = 1 = 1^2\), so \(n = 10\) works.
Next set \(\frac{n}{20-n} = 4\):
\[ n = 4(20-n) = 80 - 4n \implies 5n = 80 \implies n = 16 \]
Check it: \(\frac{16}{20-16} = \frac{16}{4} = 4 = 2^2\), so \(n = 16\) works as well.

Final Answer:
The fraction \(\frac{n}{20-n}\) turns out to be a perfect square for exactly two integers, \(n = 10\) and \(n = 16\). \[ \boxed{2} \]