2001 AMC 12 Problems/Problem 6
- 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 .
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
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.