Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(alternate solution)
Line 7: Line 7:
  
 
== Solution ==
 
== Solution ==
 +
===Solution 1===
 
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).
 
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).
  
Line 16: Line 17:
  
 
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===
 +
The last digit is 2, 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>{1\choose 2} + {3\choose 2} + {5\choose 2} + {7\choose 2} = 0 + 3 + 10 + 21 = 34</math>.
  
 
== 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}}

Revision as of 19:30, 5 January 2009

Problem

How many even three-digit integers have the property that their digits, read 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

Solution 1

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).

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 2

The last digit is 2, 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 ${1\choose 2} + {3\choose 2} + {5\choose 2} + {7\choose 2} = 0 + 3 + 10 + 21 = 34$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_9&oldid=29164"