# 2001 AMC 12 Problems/Problem 6

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## 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 $A$.

$\text{(A)}\ 4\qquad \text{(B)}\ 5\qquad \text{(C)}\ 6\qquad \text{(D)}\ 7\qquad \text{(E)}\ 8$

## Solution

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 $\text{(E)}$.

 2001 AMC 12 (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions