AMC12 2005 B

AMC12 2005 B · Q19

AMC12 2005 B · Q19. It mainly tests Linear equations, Divisibility & factors.

Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits of $x$. The integers $x$ and $y$ satisfy $x^{2}-y^{2}=m^{2}$ for some positive integer $m$. What is $x+y+m$?

设$x$和$y$为两位整数，$y$由$x$的各位数字反转得到。整数$x$和$y$满足$x^{2}-y^{2}=m^{2}$，其中$m$为某个正整数。求$x+y+m$？

(A) 88 88

(B) 112 112

(D) 144 144

(E) 154 154

Answer

Correct choice: (E)

正确答案：（E）

Solution

let $x=10a+b$, then $y=10b+a$ where $a$ and $b$ are nonzero digits. By difference of squares, \[x^2-y^2=(x+y)(x-y)\] \[=(10a+b+10b+a)(10a+b-10b-a)\] \[=(11(a+b))(9(a-b))\] For this product to be a square, the factor of $11$ must be repeated in either $(a+b)$ or $(a-b)$, and given the constraints it has to be $(a+b)=11$. The factor of $9$ is already a square and can be ignored. Now $(a-b)$ must be another square, and since $a$ cannot be $10$ or greater then $(a-b)$ must equal $4$ or $1$. If $a-b=4$ then $(a+b)+(a-b)=11+4$, $2a=15$, $a=15/2$, which is not a digit. Hence the only possible value for $a-b$ is $1$. Now we have $(a+b)+(a-b)=11+1$, $2a=12$, $a=6$, then $b=5$, $x=65$, $y=56$, $m=33$, and $x+y+m=154\Rightarrow\boxed{\mathrm{E}}$

设$x=10a+b$，则$y=10b+a$，其中$a$与$b$为非零数字。由平方差公式， \[x^2-y^2=(x+y)(x-y)\] \[=(10a+b+10b+a)(10a+b-10b-a)\] \[=(11(a+b))(9(a-b))\] 要使该乘积为完全平方，因子$11$必须在$(a+b)$或$(a-b)$中再出现一次；在给定约束下只能是$(a+b)=11$。因子$9$本身是平方，可忽略。此时$(a-b)$也必须是平方数；由于$a$不能达到$10$或更大，故$(a-b)$只能为$4$或$1$。若$a-b=4$，则$(a+b)+(a-b)=11+4$，得$2a=15$，$a=15/2$，不是数字。故唯一可能为$a-b=1$。于是$(a+b)+(a-b)=11+1$，得$2a=12$，$a=6$，从而$b=5$，$x=65$，$y=56$，$m=33$，所以$x+y+m=154\Rightarrow\boxed{\mathrm{E}}$

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