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. Also A + B cant have a carry since then for the second column, C + 1 + D cant equal D. | + | 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 D. |
Therefore <math>A+B=D</math>. | Therefore <math>A+B=D</math>. | ||
Revision as of 16:36, 20 October 2022
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. Also
cant have a carry since then for the second column,
cant equal D.
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
~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 |
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