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2001 AMC 12 Problems/Problem 6: Difference between revisions

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A cursory glance allows us to deduce that the smallest possible sum of <math>A + B + C</math> is <math>11</math> when <math>D</math>, <math>E</math>, and <math>F</math> are <math>8</math>, <math>6</math>, and <math>4</math> respectively, so this is out of the question.
A cursory glance allows us to deduce that the smallest possible sum of <math>A + B + C</math> is <math>11</math> when <math>D</math>, <math>E</math>, and <math>F</math> are <math>8</math>, <math>6</math>, and <math>4</math> respectively, so this is out of the question.


Case 2: <math>G</math>, <math>H</math>, <math>I</math>, and <math>J</math> are <math>3</math>, <math>5</math>, <math>7</math>, and <math>9</math> respectively.  
Case 2: <math>G</math>, <math>H</math>, <math>I</math>, and <math>J</math> are <math>9</math>, <math>7</math>, <math>5</math>, and <math>3</math> respectively.  


A cursory glance allows us to deduce the answer. Clearly, when <math>D</math>, <math>E</math>, and <math>F</math> are <math>6</math>, <math>4</math>, and <math>2</math> respectively, <math>A + B + C</math> is <math>9</math> when <math>A</math>, <math>B</math>, and <math>C</math> are <math>8</math>, <math>1</math>, and <math>0</math> respectively, giving us a final answer of <math>\boxed{\textbf{(E)}\ 8}</math>
A cursory glance allows us to deduce the answer. Clearly, when <math>D</math>, <math>E</math>, and <math>F</math> are <math>6</math>, <math>4</math>, and <math>2</math> respectively, <math>A + B + C</math> is <math>9</math> when <math>A</math>, <math>B</math>, and <math>C</math> are <math>8</math>, <math>1</math>, and <math>0</math> respectively, giving us a final answer of <math>\boxed{\textbf{(E)}\ 8}</math>
== Solution 2==
The ten letters must all correspond to ten distinct digits, so every digit is used in the telephone number.
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>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>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


==Video Solution by Daily Dose of Math==
==Video Solution by Daily Dose of Math==
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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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