2011 AMC 10A Problem 19

Give the squares names. Say the 1991 population was $a^2$.

1. The 2001 condition. In 2001 the population was $a^2 + 150$, and this was $9$ more than a square, say $b^2 + 9$. So
$$a^2 + 150 = b^2 + 9, \qquad\text{which means}\qquad b^2 - a^2 = 141.$$
2. Use the difference of squares. $b^2 - a^2 = (b - a)(b + a)$. So we need two whole numbers, $b - a$ (the smaller) and $b + a$ (the larger), that multiply to $141$. Since $141 = 3 \times 47$, the only choices are:
- $b - a = 1$ and $b + a = 141$. Adding the two equations: $2b = 142$, so $b = 71$ and $a = 70$.
- $b - a = 3$ and $b + a = 47$. Adding: $2b = 50$, so $b = 25$ and $a = 22$.
3. The 2011 condition decides. In 2011 the population is $a^2 + 300$, and it must be a perfect square.
- $a = 70$: $4900 + 300 = 5200$. But $72^2 = 5184$ and $73^2 = 5329$, so $5200$ is not a square. Rule it out.
- $a = 22$: $484 + 300 = 784 = 28^2$. ✓
4. So the population went from $484$ to $784$, which is $300$ more people.
5. Percent growth is the increase divided by the starting amount:
$$\frac{300}{484} \approx 0.62,$$
about $62\%$.

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

When two perfect squares differ by a known number, write the difference as $(b - a)(b + a)$ and list the ways to split that number into two factors. There are usually only a few, so each gives a candidate you can test against the other conditions. Remember that percent growth is always "increase divided by the starting value".

Start from the 2011 condition instead. If 2011 is $c^2$, then $c^2 - a^2 = 300$, so $(c - a)(c + a) = 300$. The two factors $c - a$ and $c + a$ add to $2c$, an even number, so they are both even or both odd; since $300$ is even, both are even. The even pairs are $2 \times 150$, $6 \times 50$ and $10 \times 30$, giving $a = 74$, $22$ or $10$. Now check which one makes $a^2 + 150 - 9 = a^2 + 141$ a square: $74^2 + 141 = 5617$ is not, $10^2 + 141 = 241$ is not, and $22^2 + 141 = 625 = 25^2$ is. Same answer, $a = 22$.