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Difference between revisions of "2006 AMC 12B Problems/Problem 9"

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</math>
  
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== Solution ==
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===Solution 1===
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==Solution 1 (Alcumus)==
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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==
 
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>.
  
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===Solution 2===
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==Solution 3==
 
The last digit is 4, 6, or 8.
 
The last digit is 4, 6, or 8.
  
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Thus the answer is <math>{3\choose 2} + {5\choose 2} + {7\choose 2} = 3 + 10 + 21 = \boxed{34}</math>.
 
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 3===
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==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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==Solution 5 (Forward Casework + Listing)==
 +
Casework:
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For the sake of simplicity, we are going to call the number <math>\overline{abc}</math>.
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1. If <math>a=1</math>:
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a. <math>c=2</math>. No such number exists.
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b. <math>c=4</math>. <math>b=2, 3</math>.
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c. <math>c=6</math>. <math>b=2, 3, 4, 5</math>.
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d. <math>c=8</math>. <math>b=2, 3, 4, 5, 6, 7</math>.
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2. If <math>a=2</math>: continue as above.
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We can count up that there are 34 such integers, so select <math>\boxed{B}</math>.
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 +
~hastapasta
  
 
== 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}}

Latest revision as of 16:17, 4 April 2022

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. -

Solution 5 (Forward Casework + Listing)

Casework:

For the sake of simplicity, we are going to call the number $\overline{abc}$.

1. If $a=1$:

a. $c=2$. No such number exists.

b. $c=4$. $b=2, 3$.

c. $c=6$. $b=2, 3, 4, 5$.

d. $c=8$. $b=2, 3, 4, 5, 6, 7$.

2. If $a=2$: continue as above.

We can count up that there are 34 such integers, so select $\boxed{B}$.

~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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