Difference between revisions of "2006 AMC 8 Problems/Problem 24"
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==Problem== | ==Problem== | ||
− | In the multiplication problem below <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> | + | In the multiplication problem below <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> are different digits. What is <math>A+B</math>? |
− | <cmath> \begin{ | + | <cmath> \begin{array}{cccc}& A & B & A\\ \times & & C & D\\ \hline C & D & C & D\\ \end{array} </cmath> |
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 9 </math> | <math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 9 </math> | ||
− | ==Solution== | + | ==Video Solution by OmegaLearn== |
+ | https://youtu.be/7an5wU9Q5hk?t=3080 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/sd4XopW76ps -Happytwin | ||
+ | |||
+ | https://www.youtube.com/watch?v=Y4DXkhYthhs ~David | ||
+ | |||
+ | ==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>. | <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> | ||
+ | and <math>B=0</math>. | ||
+ | |||
+ | <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 | ||
+ | \cdot 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>. | ||
+ | - J.L.L (Feel free to edit) | ||
+ | |||
+ | ==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== | ==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}} |
Latest revision as of 23:36, 17 January 2024
Contents
Problem
In the multiplication problem below , , , are different digits. What is ?
Video Solution by OmegaLearn
https://youtu.be/7an5wU9Q5hk?t=3080
Video Solution
https://youtu.be/sd4XopW76ps -Happytwin
https://www.youtube.com/watch?v=Y4DXkhYthhs ~David
Solution 1
, so . Therefore, and , so .
Solution 2
Method 1: Test
Method 2: Bash it out to time
and .
, thus the answer is
Solution 3
Because , must be . Writing it out, we can see that So, must be . . Thus, our answer is . - J.L.L (Feel free to edit)
Solution 4
We know that is 1 because after you multiply the first column and you get . Noticing that the value of does not matter as long it is a digit number, let's give the value of the digit number . After doing some multiplication using the traditional method, our product is . We know that our end product has to be , so since our value of is 10 our product should be . Therefore, is 0 because is in the spot of . We are not done as the problem is asking for the value of which is just .
- 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.