Difference between revisions of "2011 UNCO Math Contest II Problems/Problem 6"

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== Solution ==
 
== Solution ==
 
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Since all the terms past <math>5!</math> are divisible by <math>18</math>, it is only necessary to look at the remainder resulted from the first 5 terms.
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<math>1!+2!+3!+4!+5!=153\equiv 9\pmod{18}</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 23:42, 27 October 2015

Problem

What is the remainder when $1! + 2! + 3! + ?+ 2011!$ is divided by $18$?


Solution

Since all the terms past $5!$ are divisible by $18$, it is only necessary to look at the remainder resulted from the first 5 terms. $1!+2!+3!+4!+5!=153\equiv 9\pmod{18}$

See Also

2011 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions