Difference between revisions of "2001 AMC 12 Problems/Problem 6"
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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>. | ||
| − | <math>\ | + | <math>\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8</math> |
== Solution == | == Solution == | ||
Revision as of 09:45, 8 November 2021
- 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 
, where each letter represents
a different digit. The digits in each part of the number are in decreasing
order; that is, 
, 
, and 
. Furthermore,
, 
, and 
are consecutive even digits; 
, 
, 
, and 
are consecutive odd
digits; and 
. Find 
.
Solution
The last four digits 
are either 
or 
, and the other
odd digit (
or 
) must be , 
, or 
. Since , that digit must be .
Thus the sum of the two even digits in 
is 
. 
must be 
, 
, or 
, which respectively leave the pairs 
and 
, and , or and 
, as the two even digits in . Only and has sum , so is 
, and the required first digit is 
.
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. 
