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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>.
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==Solution 2==
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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 00:14, 9 February 2017

Problem

Let $S(n)$ equal the sum of the digits of positive integer $n$. For example, $S(1507) = 13$. For a particular positive integer $n$, $S(n) = 1274$. Which of the following could be the value of $S(n+1)$?

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$

Solution

Note that $n\equiv S(n)\bmod 9$, so $S(n+1)-S(n)\equiv n+1-n = 1\bmod 9$. So, since $S(n)=1274\equiv 5\bmod 9$, we have that $S(n+1)\equiv 6\bmod 9$. The only one of the answer choices $\equiv 6\bmod 9$ is $\boxed{(D)=\ 1239}$.

Solution 2

One possible value of $S(n)$ would be $1275$, but this is not any of the choices. So we know that $n$ ends in $9$, and after adding $1$, the last digit $9$ carries over. If the next digit is also a $9$, this process repeats. By the end, the sum of digits would decrease by $9$ multiplied by the number of carry-overs but increase by 1 as a result of the final carrying over. Therefore, the result must be $9x-1$ away from the original value of $S(n)$, $1274$, where $x$ is a positive integer. The only choice that satisfies this condition is $1239$, since $(1274-1239+1) \bmod 9 = 0$. So the answer is $\boxed{D}$.

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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