Difference between revisions of "2014 AMC 10B Problems/Problem 10"
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==Solution== | ==Solution== | ||
Note from the addition of the last digits that <math>A+B=D\text{ or }A+B=D+10</math>. | Note from the addition of the last digits that <math>A+B=D\text{ or }A+B=D+10</math>. | ||
| − | From the addition of the frontmost digits, <math>A+B</math> cannot have a carry, since the answer is still a five-digit number. | + | From the addition of the frontmost digits, <math>A+B</math> cannot have a carry, since the answer is still a five-digit number. Also <math>A + B</math> cant have a carry since then for the second column, <math>C + 1 + D</math> cant equal <math>D</math>. |
Therefore <math>A+B=D</math>. | Therefore <math>A+B=D</math>. | ||
Using the second or fourth column, this then implies that <math>C=0</math>, so that <math>B+C=B</math> and <math>C+D=D</math>. | Using the second or fourth column, this then implies that <math>C=0</math>, so that <math>B+C=B</math> and <math>C+D=D</math>. | ||
Note that all of the remaining equalities are now satisfied: <math>A+B=D, B+C=B,</math> and <math>B+A=D</math>. Since the digits must be distinct, the smallest possible value of <math>D</math> is <math>1+2=3</math>, and the largest possible value is <math>9</math>. Thus we have that <math>3\le D\le9</math>, so the number of possible values is <math>\boxed{\textbf{(C) }7}</math> | Note that all of the remaining equalities are now satisfied: <math>A+B=D, B+C=B,</math> and <math>B+A=D</math>. Since the digits must be distinct, the smallest possible value of <math>D</math> is <math>1+2=3</math>, and the largest possible value is <math>9</math>. Thus we have that <math>3\le D\le9</math>, so the number of possible values is <math>\boxed{\textbf{(C) }7}</math> | ||
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/x5e1DJSXFss | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | |||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 11:01, 2 July 2023
Problem
In the addition shown below 
, 
, 
, and 
are distinct digits. How many different values are possible for ?
![\[\begin{array}[t]{r} ABBCB \\ + \ BCADA \\ \hline DBDDD \end{array}\]](http://latex.artofproblemsolving.com/4/3/a/43a9d725cbf38b81b4523dbd012ce90bbd022cea.png)
Solution
Note from the addition of the last digits that 
.
From the addition of the frontmost digits, 
cannot have a carry, since the answer is still a five-digit number. Also 
cant have a carry since then for the second column, 
cant equal .
Therefore 
.
Using the second or fourth column, this then implies that 
, so that 
and 
.
Note that all of the remaining equalities are now satisfied: 
and 
. Since the digits must be distinct, the smallest possible value of is 
, and the largest possible value is 
. Thus we have that 
, so the number of possible values is 
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
| 2014 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
