Difference between revisions of "2006 AMC 12B Problems/Problem 9"
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Thus the answer is <math>{3\choose 2} + {5\choose 2} + {7\choose 2} = 3 + 10 + 21 = 34</math>. | Thus the answer is <math>{3\choose 2} + {5\choose 2} + {7\choose 2} = 3 + 10 + 21 = 34</math>. | ||
| + | |||
| + | ===Solution 3=== | ||
| + | The answer must be half of a triangular number (evens and decreasing/increasing) so <math>\boxed{34}</math> or B. | ||
| + | -zoevv | ||
== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=B|num-b=8|num-a=10}} | {{AMC12 box|year=2006|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:42, 30 December 2018
Problem
How many even three-digit integers have the property that their digits, read left to right, are in strictly increasing order?
Solution
Solution 1
Let's set the middle (tens) digit first. The middle digit can be anything from 2-7 (If it was 1 we would have the hundreds digit to be 0, if it was more than 8, the ones digit cannot be even).
If it was 2, there is 1 possibility for the hundreds digit, 3 for the ones digit. If it was 3, there are 2 possibilities for the hundreds digit, 3 for the ones digit. If it was 4, there are 3 possibilities for the hundreds digit, and 2 for the ones digit,
and so on.
So, the answer is 
.
Solution 2
The last digit is 4, 6, or 8.
If the last digit is 
, the possibilities for the first two digits correspond to 2-element subsets of 
.
Thus the answer is 
.
Solution 3
The answer must be half of a triangular number (evens and decreasing/increasing) so 
or B.
-zoevv
See also
| 2006 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
