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

Revision as of 02:19, 4 November 2024

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 $\boxed{D}$ .

Solution 4(Similar to Solution 1)

Note that a lot of numbers can have a sum of $1274$ , but what we use wishful thinking and want is some simple number $n$ where it is easy to compute the sum of the digits of $n+1$ . This number would consists of basically all digits $9$ , since when you add $1$ a lot of stuff will cancel out and end up at $0$ (ex: $399+1=400$ ). We see that the maximum number of $9$ s that can be in $1274$ is $141$ and we are left with a remainder of $5$ , so $n$ is in the form $99...9599...9$ . If we add $1$ to this number we will get $99...9600...0$ so this the sum of the digits of $n+1$ is congruent to $6 \mod 9$ . The only answer choice that is equivalent to $6 \mod 9$ is $1239$ , so our answer is $\boxed{D}$ -srisainandan6

Remark

Notice that $S(n+1)=S(n)+1-9k$ , where k is the # of carry overs that happen

~tsun26

Video Solution by OmegaLearn

https://youtu.be/zfChnbMGLVQ?t=3996

~ pi_is_3.14

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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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.

@@ Line 17: / Line 17: @@
 ==Solution 4(Similar to Solution 1)==
 Note that a lot of numbers can have a sum of <math>1274</math>, but what we use wishful thinking and want is some simple number <math>n</math> where it is easy to compute the sum of the digits of <math>n+1</math>. This number would consists of basically all digits <math>9</math>, since when you add <math>1</math> a lot of stuff will cancel out and end up at <math>0</math>(ex: <math>399+1=400</math>). We see that the maximum number of <math>9</math>s that can be in <math>1274</math> is <math>141</math> and we are left with a remainder of <math>5</math>, so <math>n</math> is in the form <math>99...9599...9</math>. If we add <math>1</math> to this number we will get <math>99...9600...0</math> so this the sum of the digits of <math>n+1</math> is congruent to <math>6 \mod 9</math>. The only answer choice that is equivalent to <math>6 \mod 9</math> is <math>1239</math>, so our answer is <math>\boxed{D}</math> -srisainandan6
+==Remark==
+Notice that <math>S(n+1)=S(n)+1-9k</math>, where k is the # of carry overs that happen
+~tsun26
 == Video Solution by OmegaLearn==

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