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

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<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>.
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==Solution 3==
 
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Because <math>DA=D</math>, <math>A</math> must be <math>1</math>. Writing it out, we can see that
 
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<math>1B1
 
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\cdot CD
 
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=0D0D
 
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+C0C0</math>
 
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So, <math>B</math> must be <math>0</math>. <math>1+0=1</math>. Thus, our answer is <math>\boxed{\textbf{(A)}\ 1}</math>.
 
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- J.L.L (Feel free to edit)
 
  
 
==Solution 4==
 
==Solution 4==

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 $A+B=1+0=\boxed{\textbf{(A)}\ 1}$.

Solution 4

We know that $A$ is 1 because after you multiply the first column $A$ and $D$ you get $D$. Noticing that the value of $CD$ does not matter as long it is a $2$ digit number, let's give the value of the $2$ digit number $CD$ $10$. After doing some multiplication using the traditional method, our product is $1B10$. We know that our end product has to be $CDCD$, so since our value of $CD$ is 10 our product should be $1010$. Therefore, $B$ is 0 because $B$ is in the spot of $0$. We are not done as the problem is asking for the value of $A+B$ which is just $\boxed{\textbf{(A)}\ 1}$.

- LearnForEver

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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