2011 AMC 10A Problems/Problem 19
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?
Let the population of the town in be . Let the population in be . It follows that . Rearrange this equation to get . Since and are both positive integers with , and also must be, and thus, they are both factors of . We have two choices for pairs of factors of : and , and and . Assuming the former pair, since must be less than , and . Solve to get . Since is not a perfect square, this is not the correct pair. Solve for the other pair to get . This time, . This is the correct pair. Now, we find the percent increase from to . Since the increase is , the percent increase is .
Proceed through the difference of squares for and :
However, instead of testing both pairs of factors we take a more certain approach. Here is the population of the town in 2011. Test through pairs of and that makes sure or . Then go through the same routine as demonstrated above to finish this problem.
Note that this approach might take more testing if one is not familiar with finding factors.
Since all the answer choices are around , we know the town's starting population must be around . We list perfect squares from to . We see that and differ by , and we can confirm that is the correct starting number by noting that . Thus, the answer is .
Let the population of the town in 1991 be and the population in 2011 be . We know that . Note that must be even. Testing, we see that and works, as , so .
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