Difference between revisions of "2006 AMC 12B Problems/Problem 9"
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− | == Solution == | + | |
− | ===Solution | + | ==Solution 1 (Alcumus)== |
+ | Let the integer have digits <math>a</math>, <math>b</math>, and <math>c</math>, read left to right. Because <math>1 \leq a<b<c</math>, none of the digits can be zero and <math>c</math> cannot be 2. If <math>c=4</math>, then <math>a</math> and <math>b</math> must each be chosen from the digits 1, 2, and 3. Therefore there are <math>\binom{3}{2}=3</math> choices for <math>a</math> and <math>b</math>, and for each choice there is one acceptable order. Similarly, for <math>c=6</math> and <math>c=8</math> there are, respectively, <math>\binom{5}{2}=10</math> and <math>\binom{7}{2}=21</math> choices for <math>a</math> and <math>b</math>. Thus there are altogether <math>3+10+21=\boxed{34}</math> such integers. | ||
+ | |||
+ | ==Solution 2== | ||
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 7, the ones digit cannot be even). | 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 7, the ones digit cannot be even). | ||
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So, the answer is <math>3(1+2)+2(3+4)+1(5+6)=\boxed{34} \Rightarrow B</math>. | So, the answer is <math>3(1+2)+2(3+4)+1(5+6)=\boxed{34} \Rightarrow B</math>. | ||
− | + | ==Solution 3== | |
The last digit is 4, 6, or 8. | The last digit is 4, 6, or 8. | ||
If the last digit is <math>x</math>, the possibilities for the first two digits correspond to 2-element subsets of <math>\{1,2,\dots,x-1\}</math>. | If the last digit is <math>x</math>, the possibilities for the first two digits correspond to 2-element subsets of <math>\{1,2,\dots,x-1\}</math>. | ||
− | 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 = \boxed{34}</math>. |
− | + | ==Solution 4== | |
The answer must be half of a triangular number (evens and decreasing/increasing) so <math>\boxed{34}</math> or the letter B. | The answer must be half of a triangular number (evens and decreasing/increasing) so <math>\boxed{34}</math> or the letter B. | ||
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Latest revision as of 18:29, 23 April 2021
Problem
How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?
Solution 1 (Alcumus)
Let the integer have digits , , and , read left to right. Because , none of the digits can be zero and cannot be 2. If , then and must each be chosen from the digits 1, 2, and 3. Therefore there are choices for and , and for each choice there is one acceptable order. Similarly, for and there are, respectively, and choices for and . Thus there are altogether such integers.
Solution 2
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 7, 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 3
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 4
The answer must be half of a triangular number (evens and decreasing/increasing) so or the letter B. -
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.