2006 AMC 8 Problems/Problem 24: Difference between revisions

Latest revision as of 13:23, 31 December 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 11: / Line 11: @@
 ==Video Solution==
-https://youtu.be/sd4XopW76ps -Happytwin
-https://www.youtube.com/channel/UCUf37EvvIHugF9gPiJ1yRzQ
 https://www.youtube.com/watch?v=Y4DXkhYthhs   ~David
-==Solution==
+==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 2==
-Method 1: Test <math>examples.</math>
-Method 2: Bash it out to <math>waste</math> time
-<math>(100A+10B+A)(10C+D) = 1000C+100D+10C+D</math>
-<math>1000AC+100BC+10AC+100AD+10BD+AD=1010C+101D</math>
-<math>1010AC+100BC+101AD = 1010C + 101D</math>
-<math>1010(A-1)(C) + 101(A-1)D + 100CB + 10BD=0</math>
-<math>A=1</math>
-,<math>B=0</math>
-And <math>0+1=1</math>, thus the answer is <math>\boxed{\textbf{(A)}\ 1}</math>
-==Solution 3==
-Because <math>DA=D</math>, <math>A</math> must be <math>1</math>. Writing it out, we can see that
-<math>1B1
-X CD
-=0D0D
-+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>.
 ==See Also==
 {{AMC8 box|year=2006|n=II|num-b=23|num-a=25}}
 {{MAA Notice}}