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

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The last four digits <math>\text{GHIJ}</math> are either <math>9753</math> or <math>7531</math>, and the other
 
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>.  
 
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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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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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 <math>\boxed{(\text{E})8}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 20:55, 16 November 2013

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 $A$.

$\text{(A)}\ 4\qquad \text{(B)}\ 5\qquad \text{(C)}\ 6\qquad \text{(D)}\ 7\qquad \text{(E)}\ 8$

Solution

The last four digits $\text{GHIJ}$ are either $9753$ or $7531$, and the other odd digit ($1$ or $9$) must be $A$, $B$, or $C$. Since $A + B + C = 9$, that digit must be $1$. Thus the sum of the two even digits in $\text{ABC}$ is $8$. $\text{DEF}$ must be $864$, $642$, or $420$, which respectively leave the pairs $2$ and $0$, $8$ and $0$, or $8$ and $6$, as the two even digits in $\text{ABC}$. Only $8$ and $0$ has sum $8$, so $\text{ABC}$ is $810$, and the required first digit is $\boxed{(\text{E})8}$.

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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