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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 (Not Rigorous)==
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==Solution 1==
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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>.
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We start with the last two terms of the polynomial <math>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}</math>, which are <math>\frac{n^5}{5!}+\frac{n^6}{6!}</math>. This can simplify to <math>\frac{6n^5+n^6}{720}</math>, which can further simplify to <math>\frac{n^5(6+n)}{720}</math>. Notice that the prime factorization of <math>720</math> is <math>5\cdot3\cdot3\cdot2\cdot2\cdot2\cdot2</math>. In order for <math>\frac{n^5(6+n)}{720}</math> to be an integer, one of the parts must divide <math>5, 3</math>, and <math>2</math>. Thus, one of the parts must be a multiple of <math>5, 3</math>, and <math>2</math>, and the LCM  of these three numbers is <math>30</math>. This means <cmath>n^5 \equiv 0\pmod{30}</cmath> or <cmath>6+n \equiv 0\pmod{30}</cmath> Thus, we can see that <math>n</math> must equal <math>0\pmod{30}</math> or <math>-6\pmod{30}</math>. Note that as long as we satisfy <math>\frac{6n^5+n^6}{720}</math>, <math>2!, 3!</math>, and <math>4!</math> will all divide into integers, as their prime factorizations will be fulfilled with the LCM being 30. E.g. <math>4! = 2\cdot2\cdot2\cdot2\cdot3</math>, and this will be divisible by <math>2^4\cdot3^4\cdot5^4</math>. Now, since we know that <math>n</math> must equal <math>0\pmod{30}</math> or <math>-6\pmod{30}</math> in order for the polynomial to be an integer, <math>n\equiv0, 24\pmod{30}</math>. To find how many integers fulfill the equation and are <math><2017</math>, we take <math>\left \lfloor\frac{2017}{30} \right \rfloor</math> and multiply it by <math>2</math>. Thus, we get <math>67\cdot2=\boxed{134}</math>.
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~Solution by IronicNinja~
  
 
==Solution 2==
 
==Solution 2==
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Note that <math>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}</math> will have a denominator that divides <math>5!</math>. Therefore, for the expression to be an integer, <math>\frac{n^6}{6!}</math> must have a denominator that divides <math>5!</math>. Thus, <math>6\mid n^6</math>, and <math>6\mid n</math>. Let <math>n=6m</math>. Substituting gives <math>1+6m+\frac{6^2m^2}{2!}+\frac{6^3m^3}{3!}+\frac{6^4m^4}{4!}+\frac{6^5m^5}{5!}+\frac{6^6m^6}{6!}</math>. Note that the first <math>5</math> terms are integers, so it suffices for <math>\frac{6^5m^5}{5!}+\frac{6^6m^6}{6!}</math> to be an integer. This simplifies to <math>\frac{6^5}{5!}m^5(m+1)=\frac{324}{5}m^5(m+1)</math>. It follows that <math>5\mid m^5(m+1)</math>. Therefore, <math>m</math> is either <math>0</math> or <math>4</math> modulo <math>5</math>. However, we seek the number of <math>n</math>, and <math>n=6m</math>. By CRT, <math>n</math> is either <math>0</math> or <math>24</math> modulo <math>30</math>, and the answer is <math>67+67=\boxed{134}</math>.
 
Note that <math>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}</math> will have a denominator that divides <math>5!</math>. Therefore, for the expression to be an integer, <math>\frac{n^6}{6!}</math> must have a denominator that divides <math>5!</math>. Thus, <math>6\mid n^6</math>, and <math>6\mid n</math>. Let <math>n=6m</math>. Substituting gives <math>1+6m+\frac{6^2m^2}{2!}+\frac{6^3m^3}{3!}+\frac{6^4m^4}{4!}+\frac{6^5m^5}{5!}+\frac{6^6m^6}{6!}</math>. Note that the first <math>5</math> terms are integers, so it suffices for <math>\frac{6^5m^5}{5!}+\frac{6^6m^6}{6!}</math> to be an integer. This simplifies to <math>\frac{6^5}{5!}m^5(m+1)=\frac{324}{5}m^5(m+1)</math>. It follows that <math>5\mid m^5(m+1)</math>. Therefore, <math>m</math> is either <math>0</math> or <math>4</math> modulo <math>5</math>. However, we seek the number of <math>n</math>, and <math>n=6m</math>. By CRT, <math>n</math> is either <math>0</math> or <math>24</math> modulo <math>30</math>, and the answer is <math>67+67=\boxed{134}</math>.
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-TheUltimate123
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==Step Solution==
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Clearly <math>1+n</math> is an integer. The part we need to verify as an integer is, upon common denominator, <math>\frac{360n^2+120n^3+30n^4+6n^5+n^6}{720}</math>. Clearly, the numerator must be even for the fraction to be an integer. Therefore, <math>n^6</math> is even and n is even, aka <math>n=2k</math> for some integer <math>k</math>. Then, we can substitute <math>n=2k</math> and see that <math>\frac{n^2}{2}</math> is trivially integral. Then, substitute for the rest of the non-confirmed-integral terms and get <math>\frac{60k^3+30k^4+12k^5+4k^6}{45}</math>. It is also clear that for this to be an integer, which it needs to be, the numerator has to be divisible by 3. The only term we worry about is the <math>4k^6</math>, and we see that <math>k=3b</math> for some integer <math>b</math>. From there we now know that <math>n=6b</math>. If we substitute again, we see that all parts except the last two fractions are trivially integral. In order for the last two fractions to sum to an integer we see that <math>n^5(6n+1) \equiv 0 \pmod5</math>, so combining with divisibility by 6, <math>n</math> is <math>24</math> or <math>0\pmod{30}</math>. There are <math>67</math> cases for each, hence the answer <math>\boxed{134}</math>.
  
 
=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}}

Latest revision as of 10:29, 18 October 2022

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

We start with the last two terms of the polynomial $1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}$, which are $\frac{n^5}{5!}+\frac{n^6}{6!}$. This can simplify to $\frac{6n^5+n^6}{720}$, which can further simplify to $\frac{n^5(6+n)}{720}$. Notice that the prime factorization of $720$ is $5\cdot3\cdot3\cdot2\cdot2\cdot2\cdot2$. In order for $\frac{n^5(6+n)}{720}$ to be an integer, one of the parts must divide $5, 3$, and $2$. Thus, one of the parts must be a multiple of $5, 3$, and $2$, and the LCM of these three numbers is $30$. This means \[n^5 \equiv 0\pmod{30}\] or \[6+n \equiv 0\pmod{30}\] Thus, we can see that $n$ must equal $0\pmod{30}$ or $-6\pmod{30}$. Note that as long as we satisfy $\frac{6n^5+n^6}{720}$, $2!, 3!$, and $4!$ will all divide into integers, as their prime factorizations will be fulfilled with the LCM being 30. E.g. $4! = 2\cdot2\cdot2\cdot2\cdot3$, and this will be divisible by $2^4\cdot3^4\cdot5^4$. Now, since we know that $n$ must equal $0\pmod{30}$ or $-6\pmod{30}$ in order for the polynomial to be an integer, $n\equiv0, 24\pmod{30}$. To find how many integers fulfill the equation and are $<2017$, we take $\left \lfloor\frac{2017}{30} \right \rfloor$ and multiply it by $2$. Thus, we get $67\cdot2=\boxed{134}$.

~Solution by IronicNinja~

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 $\frac{2010}{30}\times2=\boxed{134}$.

Solution 3

Note that $1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}$ will have a denominator that divides $5!$. Therefore, for the expression to be an integer, $\frac{n^6}{6!}$ must have a denominator that divides $5!$. Thus, $6\mid n^6$, and $6\mid n$. Let $n=6m$. Substituting gives $1+6m+\frac{6^2m^2}{2!}+\frac{6^3m^3}{3!}+\frac{6^4m^4}{4!}+\frac{6^5m^5}{5!}+\frac{6^6m^6}{6!}$. Note that the first $5$ terms are integers, so it suffices for $\frac{6^5m^5}{5!}+\frac{6^6m^6}{6!}$ to be an integer. This simplifies to $\frac{6^5}{5!}m^5(m+1)=\frac{324}{5}m^5(m+1)$. It follows that $5\mid m^5(m+1)$. Therefore, $m$ is either $0$ or $4$ modulo $5$. However, we seek the number of $n$, and $n=6m$. By CRT, $n$ is either $0$ or $24$ modulo $30$, and the answer is $67+67=\boxed{134}$.

-TheUltimate123

Step Solution

Clearly $1+n$ is an integer. The part we need to verify as an integer is, upon common denominator, $\frac{360n^2+120n^3+30n^4+6n^5+n^6}{720}$. Clearly, the numerator must be even for the fraction to be an integer. Therefore, $n^6$ is even and n is even, aka $n=2k$ for some integer $k$. Then, we can substitute $n=2k$ and see that $\frac{n^2}{2}$ is trivially integral. Then, substitute for the rest of the non-confirmed-integral terms and get $\frac{60k^3+30k^4+12k^5+4k^6}{45}$. It is also clear that for this to be an integer, which it needs to be, the numerator has to be divisible by 3. The only term we worry about is the $4k^6$, and we see that $k=3b$ for some integer $b$. From there we now know that $n=6b$. If we substitute again, we see that all parts except the last two fractions are trivially integral. In order for the last two fractions to sum to an integer we see that $n^5(6n+1) \equiv 0 \pmod5$, so combining with divisibility by 6, $n$ is $24$ or $0\pmod{30}$. There are $67$ cases for each, hence the answer $\boxed{134}$.

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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