Difference between revisions of "2017 AMC 12A Problems/Problem 18"
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==Solution== | ==Solution== | ||
| − | 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 | + | 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>. |
== See Also == | == See Also == | ||
{{AMC12 box|year=2017|ab=A|num-b=19|num-a=21}} | {{AMC12 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}} | ||
Revision as of 16:33, 8 February 2017
Problem
Let 
equal the sum of the digits of positive integer 
. For example, 
. For a particular positive integer , 
. Which of the following could be the value of 
?
Solution
Note that 
, so 
. So, since 
, we have that 
. The only one of the answer choices 
is 
.
See Also
| 2017 AMC 12A (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 12 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 | |
