Difference between revisions of "2011 UNCO Math Contest II Problems/Problem 6"
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== Problem == | == Problem == | ||
| − | What is the remainder when <math>1! + 2! + 3! + | + | What is the remainder when <math>1! + 2! + 3! + \cdots + 2011!</math> is divided by <math>18</math>? |
| − | |||
== Solution == | == Solution == | ||
| − | + | 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. | |
| + | <math>1!+2!+3!+4!+5!=153\equiv 9\pmod{18}</math> | ||
== See Also == | == See Also == | ||
Latest revision as of 23:42, 27 October 2015
Problem
What is the remainder when 
is divided by 
?
Solution
Since all the terms past 
are divisible by , it is only necessary to look at the remainder resulted from the first 5 terms.
See Also
| 2011 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| 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 | ||
