Revision as of 14:44, 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, one possibility is that $n|30$ . This yields the numbers $30,60,90,120,\cdots,2010$ . There are a total of ${67}$ numbers in the sequence. We express the last two terms as $\frac{6n^5+n^6}{720}\implies\frac{n^5(6+n)}{720}$ This yields that $n \equiv 24,30$ . Therefore, we get the final answer of $\boxed{134}$

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 ==Solution==
-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.
+The denominator contains <math>2,3,5</math>. Therefore, one possibility is that <math>n|30</math>. This yields the numbers <math>30,60,90,120,\cdots,2010</math>. There are a total of <math>{67}</math> numbers in the sequence. We express the last two terms as <math>\frac{6n^5+n^6}{720}\implies\frac{n^5(6+n)}{720}</math> This yields that <math>n \equiv 24,30</math>. Therefore, we get the final answer of <math>\boxed{134}</math>
-EDIT: 0 mod 30 and 24 mod 30 both work; therefore, the answer should be <math>\boxed{134}</math>.
 =See Also=
 {{AIME box|year=2017|n=II|num-b=7|num-a=9}}
 {{MAA Notice}}

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

Revision as of 14:44, 23 March 2017

Problem

Solution

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