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 "2009 AMC 8 Problems/Problem 22"

(→‎Solution 2 (Easy Casework))
(→‎Solution 2 (Easy Casework))
Line 15: Line 15:
 
We consider the 3 cases, where the number is 1,2 or 3 digits.
 
We consider the 3 cases, where the number is 1,2 or 3 digits.
  
−
Case 1 : The number is has <math>1</math> digit. Well, 1-9 all work except for <math>1</math>, so <math>9-1=8</math> numbers, for numbers that are 1 digit.
+
Case 1 : The number has a <math>1</math> digit. Well, 1-9 all work except for <math>1</math>, so <math>9-1=8</math> numbers, for numbers that are 1 digit.
  
 
Case 2 : The number is has <math>2</math> digits. Consider the two digit number. Since the tens digit can't be 0 or 1, we have 8 choices for the tens digit. For the ones digit, we have 9, since the ones digit can't be a 1. This gives us a total of <math>8*9=72</math> two-digit numbers.
 
Case 2 : The number is has <math>2</math> digits. Consider the two digit number. Since the tens digit can't be 0 or 1, we have 8 choices for the tens digit. For the ones digit, we have 9, since the ones digit can't be a 1. This gives us a total of <math>8*9=72</math> two-digit numbers.

Revision as of 08:17, 30 March 2020

Problem

How many whole numbers between 1 and 1000 do not contain the digit 1?

$\textbf{(A)}\ 512 \qquad \textbf{(B)}\ 648 \qquad \textbf{(C)}\ 720 \qquad \textbf{(D)}\ 728 \qquad \textbf{(E)}\ 800$

Solution

Note that this is the same as finding how many numbers with up to three digits do not contain 1.

Since there are 10 total possible digits, and only one of them is not allowed (1), each digit has its choice of 9 digits, for a total of $9*9*9=729$ such numbers. However, we over counted by one; 0 is not between 1 and 1000, so there are $\boxed{\textbf{(D)}\ 728}$ numbers.

Solution 2 (Easy Casework)

We consider the 3 cases, where the number is 1,2 or 3 digits.

Case 1 : The number has a $1$ digit. Well, 1-9 all work except for $1$, so $9-1=8$ numbers, for numbers that are 1 digit.

Case 2 : The number is has $2$ digits. Consider the two digit number. Since the tens digit can't be 0 or 1, we have 8 choices for the tens digit. For the ones digit, we have 9, since the ones digit can't be a 1. This gives us a total of $8*9=72$ two-digit numbers.

Case 3 : The number has $3$ digits. Using similar logic to Case 2, we have 8*9*9 choices for the third number.

We add the cases up getting $648+72+8$ which gives us $\framebox{(D) 728}$ numbers total.

See Also

2009 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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 AJHSME/AMC 8 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_22&oldid=120271"