2017 AMC 12A Problems/Problem 18
Contents
Problem
Let equal the sum of the digits of positive integer
. For example,
. For a particular positive integer
,
. Which of the following could be the value of
?
Solution 1
Note that , so
. So, since
, we have that
. The only one of the answer choices
is
.
Solution 2
One possible value of would be
, but this is not any of the choices. Ergo, we know that
ends in
, and after adding
, the last digit
carries over, turning the last digit into
. If the next digit is also a
, this process repeats until we get to a non-
digit. By the end, the sum of digits would decrease by
multiplied by the number of carry-overs but increase by
as a result of the final carrying over. Therefore, the result must be
less than original value of
,
, where
is a positive integer. The only choice that satisfies this condition is
, since
. The answer is
.
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 12 Problems and Solutions |
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