Difference between revisions of "2017 AMC 12A Problems/Problem 18"

Revision as of 22:32, 5 February 2018

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 1

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. Therefore, we know that $n$ ends in $9$ , and after adding $1$ , the last digit $9$ carries over, turning the last digit into $0$ . If the next digit is also a $9$ , this process repeats until we get to a non- $9$ digit. 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 $9x-1$ less than original value of $S(n)$ , $1274$ , where $x$ is a positive integer. The only choice that satisfies this condition is $\boxed{1239}$ , since $(1274-1239+1) \bmod 9 = 0$ . The answer is $\boxed{D}$ .

Solution 3

Another way to solve this is to realize that if you continuously add the digits of the number $1274 (1 + 2 + 7 + 4 = 14, 1 + 4 = 5)$ , we get $5$ . Adding one to that, we get $6$ . So, if we assess each option to see which one attains $6$ , we would discover that $1239$ satisfies the requirement, because $1 + 2 + 3 + 9 = 15$ . $1 + 5 = 6$ . The answer is $\boxed{D}$ .

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.

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 ==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. Therefore, 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>.
 ==Solution 3==

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