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 $A$ .

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

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 $\boxed{\textbf{(E) }8}$ .

2001 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 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

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2001 AMC 12 Problems/Problem 6

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