2011 AMC 10A Problem 19

Let the 1991 population be $a^2$. In 2001 it was $a^2 + 150 = b^2 + 9$, so
$$b^2 - a^2 = 141 \quad\Longrightarrow\quad (b - a)(b + a) = 141 = 3 \cdot 47.$$
The factor pairs are $1 \cdot 141$ and $3 \cdot 47$:
- $b - a = 1$, $b + a = 141$ gives $a = 70$, $b = 71$;
- $b - a = 3$, $b + a = 47$ gives $a = 22$, $b = 25$.

In 2011 the population is $a^2 + 300$, which must be a perfect square. For $a = 70$: $4900 + 300 = 5200$, not a square ($72^2 = 5184$, $73^2 = 5329$). For $a = 22$: $484 + 300 = 784 = 28^2$. So $a = 22$.

The population grew from $484$ to $784$, an increase of $300$:
$$\frac{300}{484} \approx 0.62.$$

The answer is $\boxed{\textbf{(E)}\ 62\%}$.

"Two squares differ by a known constant" is the signal for difference of squares: $(b-a)(b+a)$ must be a factor pair of the constant, and the constant $141$ has only two such pairs. Each pair gives a candidate, and the remaining condition selects the right one. Percent growth is always change over the starting value.

Skip the algebra and use the 2011 condition alone: $c^2 - a^2 = 300$ with $c - a$ and $c + a$ of the same parity, so both even: $(c - a, c + a) = (2, 150), (6, 50), (10, 30)$, giving $a = 74, 22, 10$. Then test $a^2 + 150 - 9$: only $a = 22$ gives a square ($625$). Same answer.

• Stopping at the first factorization (a = 70) without checking the 2011 condition, or misreading '9 more than a square' as 'a square plus 9 people later'.
• Computing growth as $300/784 \approx 38\%$ (dividing by the final population) instead of $300/484$.