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

(Created page with "==Problem== Let <math>S(n)</math> equal the sum of the digits of positive integer <math>n</math>. For example, <math>S(1507) = 13</math>. For a particular positive integer <ma...")
 
Line 6: Line 6:
 
==Solution==
 
==Solution==
 
Note that <math>n \equiv S(n) \pmod{9}</math>. This can be seen from the fact that <math>\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}</math>. Thus, if <math>S(n) = 1274</math>, then <math>n \equiv 5 \pmod{9}</math>, and thus <math>n+1 \equiv S(n+1) \equiv 6 \pmod{9}</math>. The only answer choice that is <math>6 \pmod{9}</math> is <math>\boxed{\textbf{(D)}\ 1265}</math>.
 
Note that <math>n \equiv S(n) \pmod{9}</math>. This can be seen from the fact that <math>\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}</math>. Thus, if <math>S(n) = 1274</math>, then <math>n \equiv 5 \pmod{9}</math>, and thus <math>n+1 \equiv S(n+1) \equiv 6 \pmod{9}</math>. The only answer choice that is <math>6 \pmod{9}</math> is <math>\boxed{\textbf{(D)}\ 1265}</math>.
 +
 +
==See Also==
 +
{{AMC10 box|year=2017|ab=A|num-b=19|num-a=21}}
 +
{{MAA Notice}}

Revision as of 18:13, 8 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 $n \equiv S(n) \pmod{9}$. 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 $\boxed{\textbf{(D)}\ 1265}$.

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png