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

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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>
 
<math>\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8</math>
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If A is odd, it must be either 9 or 1. Because it cannot be either of these values, A is even. For that reason, B and D can be eliminated. From observing trends in DEF, it can be concluded that A must be either 6 or 8. Therefore, A can be eliminated. From observing trends in GHIJ, B or C cannot be 7,5 or 3. B or C can only be 9 or 1. Because B nor C can be 9. C must be 1. If C is 1, than B can be either 0 or 2. If B is 2, then DEF must be 468. Because 6 is already taken as the value of A, then B cannot be 2. This means B is 0. This leads us to the ultimate conclusion that A is equivalent to 8.
 
  
 
== See Also ==
 
== See Also ==

Revision as of 10:36, 23 May 2023

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 $A$.

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

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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