Difference between revisions of "2014 AMC 10B Problems/Problem 10"
m |
m (→Solution) |
||
Line 17: | Line 17: | ||
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>. | + | 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>. |
− | |||
− | 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>. | ||
Any of these values can be obtained by taking <math>A=1,B=D-1</math>. | Any of these values can be obtained by taking <math>A=1,B=D-1</math>. | ||
Thus we have that <math>3\le D\le9</math>, so the number of possible values is <math>\boxed{\textbf{(C) }7}</math> | Thus we have that <math>3\le D\le9</math>, so the number of possible values is <math>\boxed{\textbf{(C) }7}</math> |
Revision as of 13:32, 12 August 2021
Contents
Problem
In the addition shown below ,
,
, and
are distinct digits. How many different values are possible for
?
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.
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
.
Any of these values can be obtained by taking
.
Thus we have that
, so the number of possible values is
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.