AMC10 2011 A

AMC10 2011 A · Q19

AMC10 2011 A · Q19. It mainly tests Perfect squares & cubes, Number theory misc.

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?

1991 年一个城镇的人口是一个完全平方数。十年后，人口增加 150 人后，比一个完全平方数多 9 人。现在，2011 年，又增加 150 人，人口再次成为一个完全平方数。以下哪一项最接近城镇在这二十年期间的人口百分比增长？

(A) 42 42

(B) 47 47

(D) 57 57

(E) 62 62

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Let \(p^2\), \(q^2+9\), and \(r^2=p^2+300\) be the populations of the town in 1991, 2001, and 2011, respectively. Then \(q^2+9=p^2+150\), so \(q^2-p^2=141\). 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 \(70^2+300=5200=52\cdot 10^2\), which is not a perfect square. Thus \(p=22\), \(p^2=484\), \(p^2+150=634=25^2+9\), and \(p^2+300=784=28^2\). The percent growth from 1991 to 2011 was \(\frac{784-484}{484}\approx 62\%\).

答案（E）：设 \(p^2\)、\(q^2+9\)、以及 \(r^2=p^2+300\) 分别为该城镇在 1991、2001 和 2011 年的人口。则 \(q^2+9=p^2+150\)，所以 \(q^2-p^2=141\)。因此 \((q-p)(q+p)=141\)，所以要么 \(q-p=3\) 且 \(q+p=47\)，要么 \(q-p=1\) 且 \(q+p=141\)。由此得到 \(p=22\) 或 \(p=70\)。注意若 \(p=70\)，则 \(70^2+300=5200=52\cdot 10^2\)，这不是完全平方数。因此 \(p=22\)，\(p^2=484\)，\(p^2+150=634=25^2+9\)，且 \(p^2+300=784=28^2\)。从 1991 到 2011 的百分比增长为 \(\frac{784-484}{484}\approx 62\%\)。

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