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

(Solution 5 (Forward Casework + Listing))
 
(15 intermediate revisions by 9 users not shown)
Line 1: Line 1:
== Problem ==
+
== 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>
{{solution}}
+
\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}}
 
{{AMC12 box|year=2006|ab=B|num-b=8|num-a=10}}
 +
{{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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png