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Difference between revisions of "2006 AMC 8 Problems/Problem 24"

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+C0C0</math>  
 
+C0C0</math>  
 
So, <math>B</math> must be <math>0</math>. <math>1+0=1</math>. Thus, our answer is <math>\boxed{\textbf{(A)}\ 1}</math>.
 
So, <math>B</math> must be <math>0</math>. <math>1+0=1</math>. Thus, our answer is <math>\boxed{\textbf{(A)}\ 1}</math>.
 +
- J.L.L (Feel free to edit)
  
 
==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}}

Revision as of 11:40, 9 September 2023

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://youtu.be/sd4XopW76ps -Happytwin

https://www.youtube.com/channel/UCUf37EvvIHugF9gPiJ1yRzQ

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

Solution

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

Solution 2

Method 1: Test $examples.$

Method 2: Bash it out to $waste$ time

$(100A+10B+A)(10C+D) = 1000C+100D+10C+D$ $1000AC+100BC+10AC+100AD+10BD+AD=1010C+101D$

$1010AC+100BC+101AD = 1010C + 101D$

$1010(A-1)(C) + 101(A-1)D + 100CB + 10BD=0$

$A=1$ ,$B=0$

And $0+1=1$, thus the answer is $\boxed{\textbf{(A)}\ 1}$

Solution 3

Because $DA=D$, $A$ must be $1$. Writing it out, we can see that $1B1 X CD =0D0D +C0C0$ So, $B$ must be $0$. $1+0=1$. Thus, our answer is $\boxed{\textbf{(A)}\ 1}$. - J.L.L (Feel free to edit)

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

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