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 "2002 AMC 12P Problems/Problem 7"

(→‎Solution)
(→‎Solution)
Line 26: Line 26:
 
This case is almost identical to the second case, just swap the <math>2</math>s with <math>3</math>s and <math>3</math>s with <math>2</math>s in the reasoning and its the same, giving us an additional <math>19</math> cases.
 
This case is almost identical to the second case, just swap the <math>2</math>s with <math>3</math>s and <math>3</math>s with <math>2</math>s in the reasoning and its the same, giving us an additional <math>19</math> cases.
  
−
Addition up all of these cases gives <math>14+19+19=52</math> cases, or <math>\boxed{\textbf{(A) 52}}.</math>
+
Addition up all of these cases gives <math>14+19+19=52</math> cases, or <math>\boxed{\textbf{(A) }52}.</math>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=6|num-a=8}}
 
{{AMC12 box|year=2002|ab=P|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:41, 30 December 2023

Problem

How many three-digit numbers have at least one $2$ and at least one $3$?

$\text{(A) }52 \qquad \text{(B) }54 \qquad \text{(C) }56 \qquad \text{(D) }58 \qquad \text{(E) }60$

Solution

We can do this problem with some simple case work.

Case 1: The hundreds place is not $2$ or $3.$ This means that the tens place and ones place must be $2$ and $3$ respectively or $3$ and $2$ respectively. This case covers $1, 4, 5, 6, 7, 8,$ and $9,$ so it gives us $2 \cdot 7 = 14$ cases.

Case 2: The hundreds place is $2.$ This means that $3$ must be in the tens place or ones place. Starting with cases in which the tens place is not $3$, we get $203, 213, 223, 243, 253, 263, 273, 283,$ and $293.$ With cases in which the tens place is $3$, we have $230-239$, or $10$ more cases. This gives us $9 + 10=19$ cases.

Case 3: The hundreds place is $3.$ This case is almost identical to the second case, just swap the $2$s with $3$s and $3$s with $2$s in the reasoning and its the same, giving us an additional $19$ cases.

Addition up all of these cases gives $14+19+19=52$ cases, or $\boxed{\textbf{(A) }52}.$

See also

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12P_Problems/Problem_7&oldid=209047"