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

Revision as of 11:38, 9 September 2023

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

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 $\boxed{\textbf{(A)}\ 1}$

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

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 40: / Line 40: @@
 ==Solution 3==
+Because <math>DA=D</math>, <math>A</math> must be 1. Writing it out, we can see that
+<math>1B1
+X CD
+=0D0D
+=C0C0</math>
+So, <math>B</math> must be 0. 1+0=1. Thus, our answer is
 ==See Also==
 {{AMC8 box|year=2006|n=II|num-b=23|num-a=25}}
 {{MAA Notice}}