AMC12 2001 A

AMC12 2001 A · Q6

AMC12 2001 A · Q6. It mainly tests Casework, Digit properties (sum of digits, divisibility tests).

A telephone number has the form $\text{ABC-DEF-GHIJ}$, where each letter represents a different digit. The digits in each part of the number are in decreasing order; that is, $A > B > C$, $D > E > F$, and $G > H > I > J$. Furthermore, $D$, $E$, and $F$ are consecutive even digits; $G$, $H$, $I$, and $J$ are consecutive odd digits; and $A + B + C = 9$. Find $A$.

电话号码的形式是 $\text{ABC-DEF-GHIJ}$，其中每个字母代表不同的数字。号码每个部分的数字是递减的，即 $A > B > C$，$D > E > F$，且 $G > H > I > J$。此外，$D$、$E$、$F$ 是连续的偶数数字；$G$、$H$、$I$、$J$ 是连续的奇数数字；并且 $A + B + C = 9$。求 $A$ 的值。

(A) 4 4

(B) 5 5

(D) 7 7

(E) 8 8

Answer

Correct choice: (E)

正确答案：（E）

Solution

We start by noting that there are $10$ letters, meaning there are $10$ digits in total. Listing them all out, we have $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$. Clearly, the most restrictive condition is the consecutive odd digits, so we create casework based on that. Case 1: $G$, $H$, $I$, and $J$ are $7$, $5$, $3$, and $1$ respectively. A cursory glance allows us to deduce that the smallest possible sum of $A + B + C$ is $11$ when $D$, $E$, and $F$ are $8$, $6$, and $4$ respectively, so this is out of the question. Case 2: $G$, $H$, $I$, and $J$ are $9$, $7$, $5$, and $3$ respectively. A cursory glance allows us to deduce the answer. Clearly, when $D$, $E$, and $F$ are $6$, $4$, and $2$ respectively, $A + B + C$ is $9$ when $A$, $B$, and $C$ are $8$, $1$, and $0$ respectively, giving us a final answer of $\boxed{\textbf{(E)}\ 8}$

我们先注意到共有 $10$ 个字母，这意味着总共用到了 $10$ 个数字。把它们全部列出为 $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$。显然，限制最强的是“连续奇数数字”的条件，因此据此分类讨论。情形 1：$G$, $H$, $I$, $J$ 分别为 $7$, $5$, $3$, $1$。粗略观察可知，当 $D$, $E$, $F$ 分别为 $8$, $6$, $4$ 时，$A + B + C$ 的最小可能和为 $11$，因此不可能满足 $A + B + C = 9$，该情形排除。情形 2：$G$, $H$, $I$, $J$ 分别为 $9$, $7$, $5$, $3$。粗略观察即可得到答案。显然，当 $D$, $E$, $F$ 分别为 $6$, $4$, $2$ 时，若 $A$, $B$, $C$ 分别为 $8$, $1$, $0$，则 $A + B + C = 9$，因此最终答案为 $\boxed{\textbf{(E)}\ 8}$。

Topics