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

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<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265</math>
 
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265</math>
  
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==Solution==
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==Solution 1==
  
 
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. 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>.
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==Solution 3==
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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 requirement, because <math>1 + 2 + 3 + 9 = 15</math>. <math>1 + 5 = 6</math>. The answer is <math>\boxed{D}</math>.
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==Solution 4(Similar to Solution 1)==
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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
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==Remark==
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Notice that <math>S(n+1)=S(n)+1-9k</math>, where <math>k</math> is the # of carry overs that happen
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~tsun26
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== Video Solution by OmegaLearn==
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https://youtu.be/zfChnbMGLVQ?t=3996
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~ pi_is_3.14
  
 
== See Also ==
 
== See Also ==
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{{AMC12 box|year=2017|ab=A|num-b=19|num-a=21}}
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{{AMC10 box|year=2017|ab=A|num-b=19|num-a=21}}
 
{{AMC12 box|year=2017|ab=A|num-b=17|num-a=19}}
 
{{AMC12 box|year=2017|ab=A|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:20, 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

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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