Difference between revisions of "2017 AMC 12A Problems/Problem 18"
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==Solution 2== | ==Solution 2== | ||
One possible value of <math>S(n)</math> would be <math>1275</math>, but this is not any of the choices. Ergo, we know that <math>n</math> ends in <math>9</math>, and after adding <math>1</math>, the last digit <math>9</math> carries over, turning the last digit into <math>0</math>. If the next digit is also a <math>9</math>, this process repeats until we get to a non-<math>9</math> digit. By the end, the sum of digits would decrease by <math>9</math> multiplied by the number of carry-overs but increase by <math>1</math> as a result of the final carrying over. Therefore, the result must be <math>9x-1</math> less than original value of <math>S(n)</math>, <math>1274</math>, where <math>x</math> is a positive integer. The only choice that satisfies this condition is <math>\boxed{1239}</math>, since <math>(1274-1239+1) \bmod 9 = 0</math>. The answer is <math>\boxed{D}</math>. | One possible value of <math>S(n)</math> would be <math>1275</math>, but this is not any of the choices. Ergo, we know that <math>n</math> ends in <math>9</math>, and after adding <math>1</math>, the last digit <math>9</math> carries over, turning the last digit into <math>0</math>. If the next digit is also a <math>9</math>, this process repeats until we get to a non-<math>9</math> digit. By the end, the sum of digits would decrease by <math>9</math> multiplied by the number of carry-overs but increase by <math>1</math> as a result of the final carrying over. Therefore, the result must be <math>9x-1</math> less than original value of <math>S(n)</math>, <math>1274</math>, where <math>x</math> is a positive integer. The only choice that satisfies this condition is <math>\boxed{1239}</math>, since <math>(1274-1239+1) \bmod 9 = 0</math>. The answer is <math>\boxed{D}</math>. | ||
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+ | ==Solution 3== | ||
+ | Another way to solve this is to realize that if you continuously add the digits of the number <math>1274 (1 + 2 + 7 + 4 = 14, 1 + 4 = 5)</math>, we get <math>5</math>. Adding one to that, we get <math>6</math>. So, if we assess each option to see which one attains <math>6</math>, we would discover that <math>1239</math> satisfies the requirements, because <math>1 + 2 + 3 + 9 = 15</math>. <math>1 + 5 = 6</math>. The answer is <math>\boxed{D}</math>. | ||
== See Also == | == See Also == |
Revision as of 19:32, 12 February 2017
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
.
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 requirements, because
.
. 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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