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

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Find the number of positive integers <math>n</math> less than <math>2017</math> such that <cmath>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}</cmath> is an integer.
 
Find the number of positive integers <math>n</math> less than <math>2017</math> such that <cmath>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}</cmath> is an integer.
  
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==Solution 1==
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==Solution 1 (Not Rigorous)==
 
Writing the last two terms with a common denominator, we have <math>\frac{6n^5+n^6}{720} \implies \frac{n^5(6+n)}{720}</math> By inspection. this yields that <math>n \equiv 0, 24 \pmod{30}</math>. Therefore, we get the final answer of <math>67 + 67 = \boxed{134}</math>.
 
Writing the last two terms with a common denominator, we have <math>\frac{6n^5+n^6}{720} \implies \frac{n^5(6+n)}{720}</math> By inspection. this yields that <math>n \equiv 0, 24 \pmod{30}</math>. Therefore, we get the final answer of <math>67 + 67 = \boxed{134}</math>.
  

Revision as of 21:41, 21 February 2018

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 1 (Not Rigorous)

Writing the last two terms with a common denominator, we have $\frac{6n^5+n^6}{720} \implies \frac{n^5(6+n)}{720}$ By inspection. this yields that $n \equiv 0, 24 \pmod{30}$. Therefore, we get the final answer of $67 + 67 = \boxed{134}$.

Solution 2

Taking out the $1+n$ part of the expression and writing the remaining terms under a common denominator, we get $\frac{1}{720}(n^6+6n^5+30n^4+120n^3+360n^2)$. Therefore the expression $n^6+6n^5+30n^4+120n^3+360n^2$ must equal $720m$ for some positive integer $m$. Taking both sides mod $2$, the result is $n^6 \equiv 0 \pmod{2}$. Therefore $n$ must be even. If $n$ is even, that means $n$ can be written in the form $2a$ where $a$ is a positive integer. Replacing $n$ with $2a$ in the expression, $64a^6+192a^5+480a^4+960a^3+1440a^2$ is divisible by $16$ because each coefficient is divisible by $16$. Therefore, if $n$ is even, $n^6+6n^5+30n^4+120n^3+360n^2$ is divisible by $16$.

Taking the equation $n^6+6n^5+30n^4+120n^3+360n^2=720m$ mod $3$, the result is $n^6 \equiv 0 \pmod{3}$. Therefore $n$ must be a multiple of $3$. If $n$ is a multiple of three, that means $n$ can be written in the form $3b$ where $b$ is a positive integer. Replacing $n$ with $3b$ in the expression, $729b^6+1458b^5+2430b^4+3240b^3+3240b^2$ is divisible by $9$ because each coefficient is divisible by $9$. Therefore, if $n$ is a multiple of $3$, $n^6+6n^5+30n^4+120n^3+360n^2$ is divisibly by $9$.

Taking the equation $n^6+6n^5+30n^4+120n^3+360n^2=720m$ mod $5$, the result is $n^6+n^5 \equiv 0 \pmod{5}$. The only values of $n (\text{mod }5)$ that satisfy the equation are $n\equiv0(\text{mod }5)$ and $n\equiv4(\text{mod }5)$. Therefore if $n$ is $0$ or $4$ mod $5$, $n^6+6n^5+30n^4+120n^3+360n^2$ will be a multiple of $5$.

The only way to get the expression $n^6+6n^5+30n^4+120n^3+360n^2$ to be divisible by $720=16 \cdot 9 \cdot 5$ is to have $n \equiv 0 \pmod{2}$, $n \equiv 0 \pmod{3}$, and $n \equiv 0 \text{ or } 4 \pmod{5}$. By the Chinese Remainder Theorem or simple guessing and checking, we see $n\equiv0,24 \pmod{30}$. Because no numbers between $2011$ and $2017$ are equivalent to $0$ or $24$ mod $30$, the answer is $\frac{2010}{30}\times2=\boxed{134}$.

See Also

2017 AIME II (Problems • Answer Key • Resources)
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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