Difference between revisions of "2017 AIME II Problems/Problem 8"

(Solution)
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==Solution==
 
==Solution==
<math>\boxed{134}</math>
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The denominator contains <math>2,3,5</math>. Therefore, <math>n|30</math>. This yields the numbers, <math>30,60,90,120,\cdots,2010</math>. There are a total of <math>\boxed{67}</math> numbers in the sequence.
  
 
=See Also=
 
=See Also=
 
{{AIME box|year=2017|n=II|num-b=7|num-a=9}}
 
{{AIME box|year=2017|n=II|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:08, 23 March 2017

Problem

Find the number of positive integers $n$ less than $2017$ such that \[1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}\] is an integer.

Solution

The denominator contains $2,3,5$. Therefore, $n|30$. This yields the numbers, $30,60,90,120,\cdots,2010$. There are a total of $\boxed{67}$ numbers in the sequence.

See Also

2017 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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