Difference between revisions of "2017 AMC 12A Problems/Problem 18"
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<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265</math> | <math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265</math> | ||
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==Solution 1== | ==Solution 1== |
Revision as of 01:38, 18 January 2021
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. Therefore, 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
.
Solution 3
Another way to solve this is to realize that if you continuously add the digits of the number , we get
. Adding one to that, we get
. So, if we assess each option to see which one attains
, we would discover that
satisfies the requirement, because
.
. The answer is
.
Solution 4(Similar to Solution 1)
Note that a lot of numbers can have a sum of , but what we use wishful thinking and want is some simple number
where it is easy to compute the sum of the digits of
. This number would consists of basically all digits
, since when you add
a lot of stuff will cancel out and end up at
(ex:
). We see that the maximum number of
s that can be in
is
and we are left with a remainder of
, so
is in the form
. If we add
to this number we will get
so this the sum of the digits of
is congruent to
. The only answer choice that is equivalent to
is
, so our answer is
-srisainandan6
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.