Question

$1991$ 年某镇人口是一个完全平方数。十年后，人口增加 $150$ 人，变成比一个完全平方数多 $9$ 。现在是 $2011$ 年，人口又增加 $150$ 人，再次成为完全平方数。下列哪一项最接近该镇这二十年人口增长的百分比？

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?

$42$

$47$

$52$

$57$

$62$

答案：E

知识点：平方差完全平方数百分数

难度评级：1750

解答：

设 $1991$ 年人口为 $p^2$ 。 $2001$ 年人口为 $q^2 + 9$ 。

于是 $p^2 + 150 = q^2 + 9$ $q^2 - p^2 = 141.$

因式分解得 $(q - p)(q + p) = 141.$

对 $p$ 、 $q$ 、 $q - p$ 、 $q + p$ 考虑，可能的因数对为 $(1, 141)$ 和 $(3, 47)$ 。

第一对满足从而得到 $q = 71$ 、 $p = 70$ 。 $q - p = 1 \text{ and } q + p = 141,$

但此时 $p^2 + 300$ 不是完全平方数，所以这一对不合要求。

检验另一对，得到 $q = 25$ 和 $p = 22$ 。

此时 $p^2 + 300 = 784$ 是完全平方数，人口增长率为 $\dfrac{300}{22^2} \cdot 100 \% \approx 62 \%.$

所以正确答案是 E。

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 $q - p = 1 \text{ and } q + p = 141,$ 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.

2011 AMC 10A 第 19 题

答案：E

难度评级：1750

解答：

其他年份的第 19 题