Question

正整数 $a$ $b$ 和 $c$ 满足 $a\ge b\ge c$ ，且以及求 $a$ 。 $a^2-b^2-c^2+ab=2011$ $a^2+3b^2+3c^2-3ab-2ac-2bc$ $=-1997.$

Let $a,$ $b,$ and $c$ be positive integers with $a\ge b\ge c$ such that $a^2-b^2-c^2+ab=2011$ and $a^2+3b^2+3c^2-3ab-2ac-2bc$ $=-1997.$ What is $a?$

$249$

$250$

$251$

$252$

$253$

答案：E

知识点：代数变形丢番图方程分类讨论

难度评级：2350

解答：

两个方程相加，得到 $2(a^2 + b^2 + c^2) - 2(ab + ac + bc)$ $= 14.$

把左边分组并因式分解，得到 $(a - b)^2 + (a - c)^2 + (b - c)^2$ $= 14.$

左边每一项都是非负整数的平方。三个平方数和为 $14$ 的唯一方式是 $9, 4$ 和 $1$ 。

在三组差中， $a - c$ 最大，所以 $a - c = 3$ 。

还不能确定另外两个差分别对应哪个平方。先试 $a - b = 1$ 和 $b - c = 2$ 。

把这些值代入第一个方程，得到化简得 $7a = 2021$ 。由于 $2021$ 不能被 $7$ 整除，所以应有 $a - b = 2$ 、 $b - c = 1$ 。 $a^2 - (a - 1)^2 - (a - 3)^2$ $+ a(a - 1) = 2011.$

再次代入并求解，得到 $b=a-2$ 。 $c=a-3$ 。 $8a-13=2011$ 。 $a=253$ 。 $\begin{aligned} a^2-(a-2)^2-(a-3)^2\\ {}+a(a-2)&=2011, \end{aligned}$

所以正确答案是 E。

Adding together the equations gives us $2(a^2 + b^2 + c^2) - 2(ab + ac + bc)$ $= 14.$

We can group terms and factor this to get $(a - b)^2 + (a - c)^2 + (b - c)^2$ $= 14.$

Note that every term on the left hand side is a nonnegative square integer. The only triple of squares that add to $14$ is $9, 4,$ and $1.$

We have that $a - c$ is the biggest difference among the three pairs. Therefore, $a - c = 3.$

We cannot discern which of the other terms we can match with the other squares. Let us try $a - b = 1$ and $b - c = 2.$

Plugging in these values into the first equation gives us $a^2 - (a - 1)^2 - (a - 3)^2$ $+ a(a - 1) = 2011.$ Simplifying yields $7a = 2021.$ Since $2021$ is not divisible by $7,$ we have that $a - b = 2$ and $b - c = 1.$

In the other case, $b=a-2$ and $c=a-3.$ The first equation becomes $\begin{aligned} a^2-(a-2)^2-(a-3)^2\\ {}+a(a-2)&=2011, \end{aligned}$ or $8a-13=2011.$ Hence $a=253.$

Thus, E is the correct answer.

2012 AMC 10A 第 24 题

答案：E

难度评级：2350

解答：

其他年份的第 24 题