# 2001 AMC 12 Problems/Problem 6

The following problem is from both the 2001 AMC 12 #6 and 2001 AMC 10 #13, so both problems redirect to this page.

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

If A is odd, it must be either 9 or 1. Because it cannot be either of these values, A is even. For that reason, B and D can be eliminated. From observing trends in DEF, it can be concluded that A must be either 6 or 8. Therefore, A can be eliminated. From observing trends in GHIJ, B or C cannot be 7,5 or 3. B or C can only be 9 or 1. Because B nor C can be 9. C must be 1. If C is 1, than B can be either 0 or 2. If B is 2, then DEF must be 468. Because 6 is already taken as the value of A, then B cannot be 2. This means B is 0. This leads us to the ultimate conclusion that A is equivalent to 8.

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