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

Revision as of 15:15, 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 $\bar{abc}$ . 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

@@ Line 31: / Line 31: @@
 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>\bar{abc}</math>.
+. 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>.
+. 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 ==
 {{AMC12 box|year=2006|ab=B|num-b=8|num-a=10}}
 {{MAA Notice}}

2006 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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