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

Latest revision as of 17:11, 8 November 2024

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

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

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

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

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

@@ Line 14: / Line 14: @@
 https://www.youtube.com/watch?v=Y4DXkhYthhs   ~David
+==Video Solution by WhyMath==
+https://youtu.be/IC_2SxI821c
 ==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>.
-==Solution 4==
-We know that <math>A</math> is 1 because after you multiply the first column <math>A</math> and <math>D</math> you get <math>D</math>. Noticing that the value of <math>CD</math> does not matter as long it is a <math>2</math> digit number, let's give the value of the <math>2</math> digit number <math>CD</math> <math>10</math>. After doing some multiplication using the traditional method, our product is <math>1B10</math>. We know that our end product has to be <math>CDCD</math>, so since our value of <math>CD</math> is 10 our product should be <math>1010</math>. Therefore, <math>B</math> is 0 because <math>B</math> is in the spot of <math>0</math>. We are not done as the problem is asking for the value of <math>A+B</math> which is just <math>\boxed{\textbf{(A)}\ 1}</math>.
-- LearnForEver
 ==See Also==
 {{AMC8 box|year=2006|n=II|num-b=23|num-a=25}}
 {{MAA Notice}}