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Difference between revisions of "2017 AMC 10A Problems/Problem 20"

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After subtracting 6 from every number, we can conclude that <math>1233</math> (originally <math>1239</math>) is the only number divisible by 9.  
 
After subtracting 6 from every number, we can conclude that <math>1233</math> (originally <math>1239</math>) is the only number divisible by 9.  
 
So our answer is <math>\boxed{\textbf{(D)}\ 1239}</math>.
 
So our answer is <math>\boxed{\textbf{(D)}\ 1239}</math>.
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~ProGameXD
 
  
 
==See Also==
 
==See Also==

Revision as of 20:16, 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 1

Note that $n \equiv S(n) \pmod{9}$. This can be seen from the fact that $\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}$. Thus, if $S(n) = 1274$, then $n \equiv 5 \pmod{9}$, and thus $n+1 \equiv S(n+1) \equiv 6 \pmod{9}$. The only answer choice that is $6 \pmod{9}$ is $\boxed{\textbf{(D)}\ 1239}$.

Solution 2

We can find out that the least number of digits the number $N$ is $142$, with $141$ $9$'s and one $5$. By randomly mixing the digits up, we are likely to get: $9999$...$9995999$...$9999$. By adding 1 to this number, we get: $9999$...$9996000$...$0000$. We can subtract 6 from every available choice, and see if the number is divisible by 9 afterwards. After subtracting 6 from every number, we can conclude that $1233$ (originally $1239$) is the only number divisible by 9. So our answer is $\boxed{\textbf{(D)}\ 1239}$.

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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