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2001 AMC 12 Problems/Problem 6: Difference between revisions

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


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


<math>3</math>, <math>5</math>, and <math>7</math> are included in both possible sequences so that we can rule out the possibilities of <math>5</math> and <math>7</math> for <math>A</math>, so <math>A</math> can only be <math>4</math>, <math>6</math>, or <math>8</math>. Since every possible sequence of DEF also contains <math>4</math>, we can rule out <math>4</math> 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 A=<math>6</math> means <math>D</math>, <math>E</math>, <math>F</math> is <math>0</math>, <math>2</math>, <math>4</math>, respectively. However, <math>6</math> and <math>8</math> must be part of <math>A</math>, <math>B</math>, or <math>C</math>, and they already sum to more than <math>10</math>, so this leaves <math>\boxed{\textbf{(E)}\ 8}</math> as 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
~megaboy6679
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{{AMC10 box|year=2001|num-b=12|num-a=14}}
{{AMC10 box|year=2001|num-b=12|num-a=14}}
{{MAA Notice}}
{{MAA Notice}}
[[Category: Introductory Algebra Problems]]

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

See Also

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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