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 8 Problems/Problem 24"

(→‎Solution 2)
(→‎Solution 4)
 
(35 intermediate revisions by 14 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
−
In the multiplication problem below <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> and  are different digits. What is <math>A+B</math>?  
+
In the multiplication problem below <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> are different digits. What is <math>A+B</math>?  
  
 
<cmath> \begin{array}{cccc}& A & B & A\\ \times & & C & D\\ \hline C & D & C & D\\ \end{array} </cmath>
 
<cmath> \begin{array}{cccc}& A & B & A\\ \times & & C & D\\ \hline C & D & C & D\\ \end{array} </cmath>
Line 7: Line 7:
 
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 9 </math>
 
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 9 </math>
  
−
==Video solution==
+
==Video Solution by OmegaLearn==
−
https://youtu.be/sd4XopW76ps -Happytwin
+
https://youtu.be/7an5wU9Q5hk?t=3080
  
−
https://youtu.be/7an5wU9Q5hk?t=3080
+
==Video Solution==
 +
https://www.youtube.com/watch?v=dQw4w9WgXcQ
  
−
https://www.youtube.com/channel/UCUf37EvvIHugF9gPiJ1yRzQ
+
https://www.youtube.com/watch?v=Y4DXkhYthhs  ~David
  
−
==Solution==
+
==Solution 1==
  
 
<math>CDCD = CD \cdot 101</math>, so <math>ABA = 101</math>. Therefore, <math>A = 1</math> and <math>B = 0</math>, so <math>A+B=1+0=\boxed{\textbf{(A)}\ 1}</math>.
 
<math>CDCD = CD \cdot 101</math>, so <math>ABA = 101</math>. Therefore, <math>A = 1</math> and <math>B = 0</math>, so <math>A+B=1+0=\boxed{\textbf{(A)}\ 1}</math>.
−
 
−
 
−
==Solution 2==
 
−
 
−
Method 1: Test examples
 
−
Method 2: When you do bash:
 
−
 
−
<math>(100A+10B+A)(10C+D) = 100C+100D+10C+D</math>
 
−
<math>100AC+100BC+100AC+100AD+10BD+AD=1010C+101D</math>
 
−
<math>1010(A-1)(C) + 101(A-1)D + 100CB+10D=0</math>
 
−
<math>A=1</math>
 
−
<math>D=0</math>
 
−
<math>B=0</math>
 
−
 
−
And now 0+1=2. I mean, 1. So the answer is <math>\boxed{\textbf{(A)}</math>
 
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2006|n=II|num-b=23|num-a=25}}
 
{{AMC8 box|year=2006|n=II|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 08:29, 18 June 2024

Problem

In the multiplication problem below $A$, $B$, $C$, $D$ are different digits. What is $A+B$?

\[\begin{array}{cccc}& A & B & A\\ \times & & C & D\\ \hline C & D & C & D\\ \end{array}\]

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 9$

Video Solution by OmegaLearn

https://youtu.be/7an5wU9Q5hk?t=3080

Video Solution

https://www.youtube.com/watch?v=dQw4w9WgXcQ

https://www.youtube.com/watch?v=Y4DXkhYthhs ~David

Solution 1

$CDCD = CD \cdot 101$, so $ABA = 101$. Therefore, $A = 1$ and $B = 0$, so $A+B=1+0=\boxed{\textbf{(A)}\ 1}$.

See Also

2006 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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=2006_AMC_8_Problems/Problem_24&oldid=220712"