Problem:
In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?
Answer Choices:
A. 42
B. 47
C. 52
D. 57
E. 62
Solution:
Let p2,q2+9, and r2=p2+300 be the populations of the town in 1991, 2001, and 2011, respectively. Then q2+9=p2+150, so q2−p2=141. Therefore (q−p)(q+p)=141, and so either q−p=3 and q+p=47, or q−p=1 and q+p=141. These give p=22 or p=70. Note that if p=70, then 702+300=5200=52⋅102, which is not a perfect square. Thus p=22, p2=484,p2+150=634=252+9, and p2+300=784=282. The percent growth from 1991 to 2011 was 484784−484​≈(E)62%​.
The problems on this page are the property of the MAA's American Mathematics Competitions