2011 AMC 10A Problem 19

Let the population in $1991$ be $p^2.$ Then let the population in $2001$ be $q^2 + 9.$ Using these values, we have $$p^2 + 150 = q^2 + 9$$ $$q^2 - p^2 = 141.$$ Factoring, we get $$(q - p)(q + p) = 141.$$ As $p$ and $q$ are integers, we have that the only possible values for $q - p$ and $q + p$ are $(1, 141)$ and $(3, 47).$ Trying the first pair, we have $$q-p=1$$ and $$q+p=141,$$ which adding together and dividing gives us $q = 71$ and $p = 70.$ We have that $p^2 + 300$ is not a square number, which means that this pair is the wrong one. Trying the other pair and using the same strategy gives us $q = 25$ and $p = 22.$ Now, $p^2 + 300 = 784,$ which is a perfect square. The percent increase in population is then $$\dfrac{300}{22^2} \cdot 100 \% \approx 62 \%.$$ Thus, E is the correct answer.