Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2001 AMC 12 Problems/Problem 6"
m (→‎Solution 2)
 
(19 intermediate revisions by 9 users not shown)
Line 1: Line 1:
−
{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #2]] and [[2001 AMC 10 Problems|2002 AMC 10A #6]]}}
+
{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #6]] and [[2001 AMC 10 Problems|2001 AMC 10 #13]]}}
  
 
== Problem ==
 
== Problem ==
Line 8: Line 8:
 
digits; and <math>A + B + C = 9</math>. Find <math>A</math>.
 
digits; and <math>A + B + C = 9</math>. Find <math>A</math>.
  
−
<math>\text{(A)}\ 4\qquad \text{(B)}\ 5\qquad \text{(C)}\ 6\qquad \text{(D)}\ 7\qquad \text{(E)}\ 8</math>
+
<math>\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8</math>
  
 
== Solution ==
 
== Solution ==
−
The last four digits <math>\text{GHIJ}</math> are either <math>9753</math> or <math>7531</math>, and the other
+
 
−
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>.
+
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.
−
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>.
+
 
 +
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.
 +
 
 +
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.
 +
 
 +
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 <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>.
 +
 
 +
<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.
 +
 
 +
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.
 +
 
 +
~megaboy6679
 +
 
 +
==Video Solution by Daily Dose of Math==
 +
 
 +
https://youtu.be/z7o_BiWLDlk?si=9aZ0zIx2lkh_8CV3
 +
 
 +
~Thesmartgreekmathdude
  
 
== See Also ==
 
== See Also ==
Line 19: Line 44:
 
{{AMC12 box|year=2001|num-b=5|num-a=7}}
 
{{AMC12 box|year=2001|num-b=5|num-a=7}}
 
{{AMC10 box|year=2001|num-b=12|num-a=14}}
 
{{AMC10 box|year=2001|num-b=12|num-a=14}}
 +
{{MAA Notice}}

Latest revision as of 13:02, 12 October 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 $\boxed{\textbf{(E)}\ 8}$

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_6&oldid=229670"