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, read left to right, are in strictly increasing order?  
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How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?  
  
 
<math>
 
<math>
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</math>
 
</math>
  
== Solution ==
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===Solution 1===
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==Solution 1 (Alcumus)==
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).
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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.
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==Solution 2==
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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 2, there is 1 possibility for the hundreds digit, 3 for the ones digit.
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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 2===
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==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>.
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Thus the answer is <math>{3\choose 2} + {5\choose 2} + {7\choose 2} = 3 + 10 + 21 = \boxed{34}</math>.
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==Solution 4==
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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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== 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}}
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{{MAA Notice}}

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?

$\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150$


Solution 1 (Alcumus)

Let the integer have digits $a$, $b$, and $c$, read left to right. Because $1 \leq a<b<c$, none of the digits can be zero and $c$ cannot be 2. If $c=4$, then $a$ and $b$ must each be chosen from the digits 1, 2, and 3. Therefore there are $\binom{3}{2}=3$ choices for $a$ and $b$, and for each choice there is one acceptable order. Similarly, for $c=6$ and $c=8$ there are, respectively, $\binom{5}{2}=10$ and $\binom{7}{2}=21$ choices for $a$ and $b$. Thus there are altogether $3+10+21=\boxed{34}$ 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 $3(1+2)+2(3+4)+1(5+6)=\boxed{34} \Rightarrow B$.

Solution 3

The last digit is 4, 6, or 8.

If the last digit is $x$, the possibilities for the first two digits correspond to 2-element subsets of $\{1,2,\dots,x-1\}$.

Thus the answer is ${3\choose 2} + {5\choose 2} + {7\choose 2} = 3 + 10 + 21 = \boxed{34}$.

Solution 4

The answer must be half of a triangular number (evens and decreasing/increasing) so $\boxed{34}$ or the letter B. -

See also

2006 AMC 12B (ProblemsAnswer KeyResources)
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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