2001 AMC 12 Problems/Problem 6

Problem

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 $A$ .

The last four digits $\text{GHIJ}$ are either $9753$ or $7531$ , and the other odd digit ( $1$ or $9$ ) must be $A$ , $B$ , or $C$ . Since $A + B + C = 9$ , that digit must be $1$ . Thus the sum of the two even digits in $\text{ABC}$ is $8$ . $\text{DEF}$ must be $864$ , $642$ , or $420$ , which respectively leave the pairs $2$ and $0$ , $8$ and $0$ , or $8$ and $6$ , as the two even digits in $\text{ABC}$ . Only $8$ and $0$ has sum $8$ , so $\text{ABC}$ is $810$ , and the required first digit is 8, so the answer is $\text{(E)}$ .