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Difference between revisions of "2001 AMC 12 Problems/Problem 6"

(New page: == Problem == A telephone number has the form <math>\text{ABC-DEF-GHIJ}</math>, where each letter represents a different digit. The digits in each part of the number are in decreasing orde...)
 
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{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #6]] and [[2001 AMC 10 Problems|2001 AMC 10 #13]]}}
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== Problem ==
 
== Problem ==
 
A telephone number has the form <math>\text{ABC-DEF-GHIJ}</math>, where each letter represents
 
A telephone number has the form <math>\text{ABC-DEF-GHIJ}</math>, where each letter represents
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digits; and <math>A + B + C = 9</math>. Find <math>A</math>.
 
digits; and <math>A + B + C = 9</math>. Find <math>A</math>.
  
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<math>\text{(A)}\ 4\qquad \text{(B)}\ 5\qquad \text{(C)}\ 6\qquad \text{(D)}\ 7\qquad \text{(E)}\ 8</math>
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<math>\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8</math>
  
 
== Solution ==
 
== Solution ==
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The last four digits <math>\text{GHIJ}</math> are either <math>9753</math> or <math>7531</math>, and the other
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odd digit (<math>1</math> or <math>9</math>) must be <math>A</math>, <math>B</math>, or <math>C</math>. Since <math>A + B + C = 9</math>, that digit must be <math>1</math>.  
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We start by noting that there are <math>10</math> letters, meaning there are <math>10</math> digits in total. Listing them all out, we have <math>0, 1, 2, 3, 4, 5, 6, 7, 8, 9</math>. Clearly, the most restrictive condition is the consecutive odd digits, so we create casework based on that.
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Thus the sum of the two even digits in <math>\text{ABC}</math> is <math>8</math>. <math>\text{DEF}</math> must be <math>864</math>, <math>642</math>, or <math>420</math>, which respectively leave the pairs <math>2</math> and <math>0</math>, <math>8</math> and <math>0</math>, or <math>8</math> and <math>6</math>, as the two even digits in <math>\text{ABC}</math>. Only <math>8</math> and <math>0</math> has sum <math>8</math>, so <math>\text{ABC}</math> is <math>810</math>, and the required first digit is 8, so the answer is <math>\text{(E)}</math>.
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Case 1: <math>G</math>, <math>H</math>, <math>I</math>, and <math>J</math> are <math>7</math>, <math>5</math>, <math>3</math>, and <math>1</math> respectively.  
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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.
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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.
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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>
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==Video Solution by Daily Dose of Math==
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https://youtu.be/z7o_BiWLDlk?si=9aZ0zIx2lkh_8CV3
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~Thesmartgreekmathdude
  
 
== See Also ==
 
== See Also ==
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{{AMC12 box|year=2001|num-b=5|num-a=7}}
 
{{AMC12 box|year=2001|num-b=5|num-a=7}}
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{{AMC10 box|year=2001|num-b=12|num-a=14}}
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{{MAA Notice}}

Revision as of 20:43, 15 July 2024

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

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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