Difference between revisions of "2006 AMC 12B Problems/Problem 9"
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− | == Problem == | + | == 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 == | + | <math> |
+ | \text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150 | ||
+ | </math> | ||
+ | |||
+ | |||
+ | ==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). | ||
+ | |||
+ | 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 <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. | ||
+ | |||
+ | 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 = \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. | ||
+ | - | ||
+ | |||
+ | ==Solution 5 (Forward Casework + Listing)== | ||
+ | Casework: | ||
+ | |||
+ | For the sake of simplicity, we are going to call the number <math>\overline{abc}</math>. | ||
+ | |||
+ | 1. If <math>a=1</math>: | ||
+ | |||
+ | a. <math>c=2</math>. No such number exists. | ||
+ | |||
+ | b. <math>c=4</math>. <math>b=2, 3</math>. | ||
+ | |||
+ | c. <math>c=6</math>. <math>b=2, 3, 4, 5</math>. | ||
+ | |||
+ | d. <math>c=8</math>. <math>b=2, 3, 4, 5, 6, 7</math>. | ||
+ | |||
+ | 2. If <math>a=2</math>: continue as above. | ||
+ | |||
+ | We can count up that there are 34 such integers, so select <math>\boxed{B}</math>. | ||
+ | |||
+ | ~hastapasta | ||
== See also == | == See also == | ||
− | + | {{AMC12 box|year=2006|ab=B|num-b=8|num-a=10}} | |
+ | {{MAA Notice}} |
Latest revision as of 15:17, 4 April 2022
Contents
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. -
Solution 5 (Forward Casework + Listing)
Casework:
For the sake of simplicity, we are going to call the number .
1. If :
a. . No such number exists.
b. . .
c. . .
d. . .
2. If : continue as above.
We can count up that there are 34 such integers, so select .
~hastapasta
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 |
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