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: Rewrite the expression using a new variable.
Let $m = 20 - n$, so $n = 20 - m$ and the given expression becomes $\dfrac{20-m}{m} = \dfrac{20}{m} - 1$. For this to be the square of an integer, $\dfrac{20}{m}$ must be a whole number and $\dfrac{20}{m} - 1$ must be a perfect square.

Step 2: Check divisors of 20 that keep n in a sensible range.
Since $m$ has to divide 20 exactly, look at the divisors of 20 and see which ones make $\dfrac{20}{m} - 1$ a perfect square.
Take $m = 4$:
\[ \frac{20}{4} - 1 = 5 - 1 = 4 = 2^2 \]
This is a perfect square, so it works. Here $n = 20 - m = 16$.
Take $m = 10$:
\[ \frac{20}{10} - 1 = 2 - 1 = 1 = 1^2 \]
This is also a perfect square, so it works too. Here $n = 20 - m = 10$.

Step 3: Confirm both values in the original expression.
For $n = 16$: $\dfrac{16}{20-16} = \dfrac{16}{4} = 4$, the square of 2.
For $n = 10$: $\dfrac{10}{20-10} = \dfrac{10}{10} = 1$, the square of 1.

Final Answer:
There are exactly two integers, $n = 10$ and $n = 16$, for which the expression is the square of an integer. \[ \boxed{2} \]