AMC12 2007 B

AMC12 2007 B · Q18

AMC12 2007 B · Q18. It mainly tests Linear equations, Money / coins.

Let $a$, $b$, and $c$ be digits with $a\ne 0$. The three-digit integer $abc$ lies one third of the way from the square of a positive integer to the square of the next larger integer. The integer $acb$ lies two thirds of the way between the same two squares. What is $a+b+c$?

设$a$、$b$、$c$为数字且$a\ne 0$。三位整数$abc$位于某个正整数的平方与下一个更大整数的平方之间的三分之一处。整数$acb$位于同一对平方之间的三分之二处。求$a+b+c$。

(A) 10 10

(B) 13 13

(D) 18 18

(E) 21 21

Answer

Correct choice: (C)

正确答案：（C）

Solution

The difference between $acb$ and $abc$ is given by $(100a + 10c + b) - (100a + 10b + c) = 9(c-b)$ The difference between the two squares is three times this amount or $27(c-b)$ The difference between two consecutive squares is always an odd number, therefore $c-b$ is odd. We will show that $c-b$ must be 1. Otherwise we would be looking for two consecutive squares that are at least 81 apart. But already the equation $(x+1)^2-x^2 = 27\cdot 3$ solves to $x=40$, and $40^2$ has more than three digits. The consecutive squares with common difference $27$ are $13^2=169$ and $14^2=196$. One third of the way between them is $178$ and two thirds of the way is $187$. This gives $a=1$, $b=7$, $c=8$. $a+b+c = 16 \Rightarrow \mathrm{(C)}$

$acb$与$abc$的差为 $(100a + 10c + b) - (100a + 10b + c) = 9(c-b)$ 两平方之差是这个差的三倍，即 $27(c-b)$ 相邻两个平方的差总是奇数，因此$c-b$为奇数。我们将证明$c-b$必须为1。否则我们需要寻找一对相邻平方，它们至少相差81。但方程$(x+1)^2-x^2 = 27\cdot 3$解得$x=40$，而$40^2$已超过三位数。公差为$27$的相邻平方是$13^2=169$与$14^2=196$。它们之间的三分之一处是$178$，三分之二处是$187$。因此$a=1$, $b=7$, $c=8$。 $a+b+c = 16 \Rightarrow \mathrm{(C)}$

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