Difference between revisions of "2017 AMC 12A Problems/Problem 18"
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Note that <math>n\equiv S(n)\bmod 9</math>, so <math>S(n+1)-S(n)\equiv n+1-n = 1\bmod 9</math>. So, since <math>S(n)=1274\equiv 5\bmod 9</math>, we have that <math>S(n+1)\equiv 6\bmod 9</math>. The only one of the answer choices <math>\equiv 6\bmod 9</math> is <math>\boxed{(D)=\ 1239}</math>. | Note that <math>n\equiv S(n)\bmod 9</math>, so <math>S(n+1)-S(n)\equiv n+1-n = 1\bmod 9</math>. So, since <math>S(n)=1274\equiv 5\bmod 9</math>, we have that <math>S(n+1)\equiv 6\bmod 9</math>. The only one of the answer choices <math>\equiv 6\bmod 9</math> is <math>\boxed{(D)=\ 1239}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | One possible value of <math>S(n)</math> would be <math>1275</math>, but this is not any of the choices. So 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. If the next digit is also a <math>9</math>, this process repeats. By the end, the sum of digits would decrease by <math>9</math> multiplied by the number of carry-overs but increase by 1 as a result of the final carrying over. Therefore, the result must be <math>9x-1</math> away from the 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>1239</math>, since <math>(1274-1239+1) \bmod 9 = 0</math>. So the answer is <math>\boxed{D}</math>. | ||
== See Also == | == See Also == |
Revision as of 01:14, 9 February 2017
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
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. So we know that
ends in
, and after adding
, the last digit
carries over. If the next digit is also a
, this process repeats. By the end, the sum of digits would decrease by
multiplied by the number of carry-overs but increase by 1 as a result of the final carrying over. Therefore, the result must be
away from the original value of
,
, where
is a positive integer. The only choice that satisfies this condition is
, since
. So the answer is
.
See Also
2017 AMC 12A (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 12 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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