Difference between revisions of "2014 AMC 10B Problems/Problem 10"
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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> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/CCOjtLn2AKM | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=9|num-a=11}} | {{AMC10 box|year=2014|ab=B|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:24, 7 August 2020
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
.
Thus, if we have any
such that
then the addition will be satisfied.
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.