2001 AMC 12 Problems/Problem 6: Difference between revisions

Latest revision as of 22:36, 11 November 2025

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 $9$ , $7$ , $5$ , and $3$ 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 $3$ , $5$ , and $7$ for $A$ , so $A$ can only be $4$ , $6$ , or $8$ (Note that we can also rule out $9$ as a possibility since there is no set of $2$ unique digits that could sum with $9$ to $9$ ). 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 then be part of $ABC$ , and they already sum to more than $9$ . So then we can try $A$ = $8$ . We have two cases: Where $DEF$ includes $0$ , $2$ , and $4$ , or when $DEF$ includes $2$ , $4$ , and $6$ . If we say the former, $ABC$ includes $6$ as well as $8$ , which, again, has a sum over $9$ . Therefore, $DEF$ must include $2$ , $4$ , and $6$ , and leaves us with $0$ , $1$ , and $8$ for $ABC$ . This sums to 9, and since the largest digit is $8$ , $\boxed{\textbf{(E)}\ 8}$ is our answer.

~megaboy6679

Video Solution by Daily Dose of Math

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

~Thesmartgreekmathdude

@@ Line 28: / Line 28: @@
 We note that <imath>1</imath>, <imath>3</imath>, <imath>5</imath>, <imath>7</imath>, <imath>9</imath> are the odd numbers, and possible sequences for <imath>GHIJ</imath> are either <imath>1</imath>, <imath>3</imath>, <imath>5</imath>, <imath>7</imath> or <imath>3</imath>, <imath>5</imath>, <imath>7</imath>, <imath>9</imath>.
-<imath>3</imath>, <imath>5</imath>, and <imath>7</imath> are included in both possible sequences so that we can rule out the possibilities of <imath>3</imath>, <imath>5</imath>, and <imath>7</imath> for <imath>A</imath>, so <imath>A</imath> can only be <imath>4</imath>, <imath>6</imath>, or <imath>8</imath> (Note that we can also rule out <imath>9</imath> as a possibility since there is no set of <imath>2</imath> unique digits that could sum with <imath>9</imath> to <imath>9</imath>). Since every possible sequence of <imath>D</imath>, <imath>E</imath>, <imath>F</imath> also contains <imath>4</imath>, we can rule out <imath>4</imath> as well.
+<imath>3</imath>, <imath>5</imath>, and <imath>7</imath> are included in both possible sequences so that we can rule out the possibilities of <imath>3</imath>, <imath>5</imath>, and <imath>7</imath> for <imath>A</imath>, so <imath>A</imath> can only be <imath>4</imath>, <imath>6</imath>, or <imath>8</imath> (Note that we can also rule out <imath>9</imath> as a possibility since there is no set of <imath>2</imath> unique digits that could sum with <imath>9</imath> to <imath>9</imath>). Since every possible sequence of <imath>DEF</imath> also contains <imath>4</imath>, we can rule out <imath>4</imath> as well.
-Testing <imath>A</imath>=<imath>6</imath> means <imath>D</imath>, <imath>E</imath>, <imath>F</imath> is <imath>0</imath>, <imath>2</imath>, <imath>4</imath>, respectively. However, <imath>6</imath> and <imath>8</imath> must then be part of <imath>A</imath>, <imath>B</imath>, and <imath>C</imath>, and they already sum to more than <imath>9</imath>. So then we can try <imath>A</imath>=<imath>8</imath>. We have two cases: Where <imath>D</imath>, <imath>E</imath>, <imath>F</imath> includes <imath>0</imath>, <imath>2</imath>, and <imath>4</imath>, or when <imath>D</imath>, <imath>E</imath>, <imath>F</imath> includes <imath>2</imath>, <imath>4</imath>, and <imath>6</imath>. If we say the former, <imath>A</imath>, <imath>B</imath>, <imath>C</imath> includes <imath>6</imath> as well as <imath>8</imath>, which, again, has a sum over <imath>9</imath>. Therefore, <imath>D</imath>, <imath>E</imath>, <imath>F</imath> must include <imath>2</imath>, <imath>4</imath>, and <imath>6</imath>, leaves us with <imath>0</imath>, <imath>1</imath>, and <imath>8</imath> for <imath>A</imath>, <imath>B</imath>, <imath>C</imath>. This sums to 9, and since the largest digit is <imath>8</imath>, <imath>\boxed{\textbf{(E)}\ 8}</imath> is our answer.
+Testing <imath>A</imath>=<imath>6</imath> means <imath>D</imath>, <imath>E</imath>, <imath>F</imath> is <imath>0</imath>, <imath>2</imath>, <imath>4</imath>, respectively. However, <imath>6</imath> and <imath>8</imath> must then be part of <imath>ABC</imath>, and they already sum to more than <imath>9</imath>. So then we can try <imath>A</imath>=<imath>8</imath>. We have two cases: Where <imath>DEF</imath> includes <imath>0</imath>, <imath>2</imath>, and <imath>4</imath>, or when <imath>DEF</imath> includes <imath>2</imath>, <imath>4</imath>, and <imath>6</imath>. If we say the former, <imath>ABC</imath> includes <imath>6</imath> as well as <imath>8</imath>, which, again, has a sum over <imath>9</imath>. Therefore, <imath>DEF</imath> must include <imath>2</imath>, <imath>4</imath>, and <imath>6</imath>, and leaves us with <imath>0</imath>, <imath>1</imath>, and <imath>8</imath> for <imath>ABC</imath>. This sums to 9, and since the largest digit is <imath>8</imath>, <imath>\boxed{\textbf{(E)}\ 8}</imath> is our answer.
 ~megaboy6679

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

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