AMC10 2008 B

AMC10 2008 B · Q23

AMC10 2008 B · Q23. It mainly tests Linear equations, Divisibility & factors.

A rectangular floor measures $a$ feet by $b$ feet, where $a$ and $b$ are positive integers with $b > a$. An artist paints a rectangle on the floor with the sides of the rectangle parallel to the sides of the floor. The unpainted part of the floor forms a border of width 1 foot around the painted rectangle and occupies half the area of the entire floor. How many possibilities are there for the ordered pair $(a, b)$?

一个矩形地板尺寸为$a$英尺乘$b$英尺，其中$a$和$b$是正整数且$b>a$。艺术家在地板上画一个矩形，其边与地板边平行。未涂部分形成宽度1英尺的边框，占据整个地板面积的一半。有多少种可能的有序对$(a,b)$？

(A) 1 1

(B) 2 2

(D) 4 4

(E) 5 5

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Because the area of the border is half the area of the floor, the same is true of the painted rectangle. The painted rectangle measures $a-2$ by $b-2$ feet. Hence $ab=2(a-2)(b-2)$, from which $0=ab-4a-4b+8$. Add 8 to each side of the equation to produce $$8=ab-4a-4b+16=(a-4)(b-4).$$ Because the only integer factorizations of 8 are $$8=1\cdot 8=2\cdot 4=(-4)\cdot(-2)=(-8)\cdot(-1),$$ and because $b>a>0$, the only possible ordered pairs satisfying this equation for $(a-4,b-4)$ are $(1,8)$ and $(2,4)$. Hence $(a,b)$ must be one of the two ordered pairs $(5,12)$, or $(6,8)$.

答案（B）：因为边框的面积是地板面积的一半，所以涂色矩形的面积也同样如此。涂色矩形的尺寸为 $(a-2)$ 英尺乘 $(b-2)$ 英尺。因此 $ab=2(a-2)(b-2)$，从而得到 $0=ab-4a-4b+8$。在等式两边都加上 8，得到 $$8=ab-4a-4b+16=(a-4)(b-4).$$ 因为 8 的整数分解只有 $$8=1\cdot 8=2\cdot 4=(-4)\cdot(-2)=(-8)\cdot(-1),$$ 并且由于 $b>a>0$，使该等式对 $(a-4,b-4)$ 成立的有序对只能是 $(1,8)$ 和 $(2,4)$。因此 $(a,b)$ 必须是两个有序对之一：$(5,12)$ 或 $(6,8)$。

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