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

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 $3$ , $5$ , $7$ , and $9$ 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 $\boxed{\textbf{(E)}\ 8}$

Solution 2

The ten letters must all correspond to ten distinct digits, so every digit is used in the telephone number.

We note that $1$ , $3$ , $5$ , $7$ , $9$ are the odd numbers, and possible sequences for $GHIJ$ are either $1$ , $3$ , $5$ , $7$ or $3$ , $5$ , $7$ , $9$ .

$3$ , $5$ , and $7$ are included in both possible sequences so that we can rule out the possibilities of $5$ and $7$ for $A$ , so $A$ can only be $4$ , $6$ , or $8$ . Since every possible sequence of DEF also contains $4$ , we can rule out $4$ as well.

Testing A= $6$ means $D$ , $E$ , $F$ is $0$ , $2$ , $4$ , respectively. However, $6$ and $8$ must be part of $A$ , $B$ , or $C$ , and they already sum to more than $10$ , so this leaves $\boxed{\textbf{(E)}\ 8}$ as our answer.

~megaboy6679

Video Solution by Daily Dose of Math

https://youtu.be/z7o_BiWLDlk?si=9aZ0zIx2lkh_8CV3

~Thesmartgreekmathdude

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

Contents

Problem

Solution

Solution 2

Video Solution by Daily Dose of Math

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