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. | ||
− | ==Solution 1 (Not Rigorous)== | + | ==Solution 1== |
+ | 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\cdot1</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>0\pmod{30}-6</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>0\pmod{30}-6</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>\lfloor\frac{2017}{30}\rfloor</math> and multiply it by <math>2</math>. Thus, we get <math>67\cdot2=\boxed{134}</math>. | ||
+ | |||
+ | Solution by IronicNinja~ | ||
+ | |||
+ | ==Solution 2 (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>. | ||
− | ==Solution | + | ==Solution 3== |
Taking out the <math>1+n</math> part of the expression and writing the remaining terms under a common denominator, we get <math>\frac{1}{720}(n^6+6n^5+30n^4+120n^3+360n^2)</math>. Therefore the expression <math>n^6+6n^5+30n^4+120n^3+360n^2</math> must equal <math>720m</math> for some positive integer <math>m</math>. | Taking out the <math>1+n</math> part of the expression and writing the remaining terms under a common denominator, we get <math>\frac{1}{720}(n^6+6n^5+30n^4+120n^3+360n^2)</math>. Therefore the expression <math>n^6+6n^5+30n^4+120n^3+360n^2</math> must equal <math>720m</math> for some positive integer <math>m</math>. | ||
Taking both sides mod <math>2</math>, the result is <math>n^6 \equiv 0 \pmod{2}</math>. Therefore <math>n</math> must be even. If <math>n</math> is even, that means <math>n</math> can be written in the form <math>2a</math> where <math>a</math> is a positive integer. Replacing <math>n</math> with <math>2a</math> in the expression, <math>64a^6+192a^5+480a^4+960a^3+1440a^2</math> is divisible by <math>16</math> because each coefficient is divisible by <math>16</math>. Therefore, if <math>n</math> is even, <math>n^6+6n^5+30n^4+120n^3+360n^2</math> is divisible by <math>16</math>. | Taking both sides mod <math>2</math>, the result is <math>n^6 \equiv 0 \pmod{2}</math>. Therefore <math>n</math> must be even. If <math>n</math> is even, that means <math>n</math> can be written in the form <math>2a</math> where <math>a</math> is a positive integer. Replacing <math>n</math> with <math>2a</math> in the expression, <math>64a^6+192a^5+480a^4+960a^3+1440a^2</math> is divisible by <math>16</math> because each coefficient is divisible by <math>16</math>. Therefore, if <math>n</math> is even, <math>n^6+6n^5+30n^4+120n^3+360n^2</math> is divisible by <math>16</math>. | ||
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The only way to get the expression <math>n^6+6n^5+30n^4+120n^3+360n^2</math> to be divisible by <math>720=16 \cdot 9 \cdot 5</math> is to have <math>n \equiv 0 \pmod{2}</math>, <math>n \equiv 0 \pmod{3}</math>, and <math>n \equiv 0 \text{ or } 4 \pmod{5}</math>. By the Chinese Remainder Theorem or simple guessing and checking, we see <math>n\equiv0,24 \pmod{30}</math>. Because no numbers between <math>2011</math> and <math>2017</math> are equivalent to <math>0</math> or <math>24</math> mod <math>30</math>, the answer is <math>\frac{2010}{30}\times2=\boxed{134}</math>. | The only way to get the expression <math>n^6+6n^5+30n^4+120n^3+360n^2</math> to be divisible by <math>720=16 \cdot 9 \cdot 5</math> is to have <math>n \equiv 0 \pmod{2}</math>, <math>n \equiv 0 \pmod{3}</math>, and <math>n \equiv 0 \text{ or } 4 \pmod{5}</math>. By the Chinese Remainder Theorem or simple guessing and checking, we see <math>n\equiv0,24 \pmod{30}</math>. Because no numbers between <math>2011</math> and <math>2017</math> are equivalent to <math>0</math> or <math>24</math> mod <math>30</math>, the answer is <math>\frac{2010}{30}\times2=\boxed{134}</math>. | ||
− | ==Solution | + | ==Solution 4== |
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>. |
Revision as of 10:40, 2 June 2018
Contents
Problem
Find the number of positive integers less than
such that
is an integer.
Solution 1
We start with the last two terms of the polynomial , which are
. This can simplify to
, which can further simplify to
. Notice that the prime factorization of
is
. In order for
to be an integer, one of the parts must divide
, and
. Thus, one of the parts must be a multiple of
, and
, and the LCM of these three numbers is
. This means
or
Thus, we can see that
must equal
or
. Note that as long as we satisfy
,
, and
will all divide into integers, as their prime factorizations will be fulfilled with the LCM being 30. E.g.
, and this will be divisible by
. Now, since we know that
must equal
or
in order for the polynomial to be an integer,
. To find how many integers fulfill the equation and are
, we take
and multiply it by
. Thus, we get
.
Solution by IronicNinja~
Solution 2 (Not Rigorous)
Writing the last two terms with a common denominator, we have By inspection. this yields that
. Therefore, we get the final answer of
.
Solution 3
Taking out the part of the expression and writing the remaining terms under a common denominator, we get
. Therefore the expression
must equal
for some positive integer
.
Taking both sides mod
, the result is
. Therefore
must be even. If
is even, that means
can be written in the form
where
is a positive integer. Replacing
with
in the expression,
is divisible by
because each coefficient is divisible by
. Therefore, if
is even,
is divisible by
.
Taking the equation mod
, the result is
. Therefore
must be a multiple of
. If
is a multiple of three, that means
can be written in the form
where
is a positive integer. Replacing
with
in the expression,
is divisible by
because each coefficient is divisible by
. Therefore, if
is a multiple of
,
is divisibly by
.
Taking the equation mod
, the result is
. The only values of
that satisfy the equation are
and
. Therefore if
is
or
mod
,
will be a multiple of
.
The only way to get the expression to be divisible by
is to have
,
, and
. By the Chinese Remainder Theorem or simple guessing and checking, we see
. Because no numbers between
and
are equivalent to
or
mod
, the answer is
.
Solution 4
Note that will have a denominator that divides
. Therefore, for the expression to be an integer,
must have a denominator that divides
. Thus,
, and
. Let
. Substituting gives
. Note that the first
terms are integers, so it suffices for
to be an integer. This simplifies to
. It follows that
. Therefore,
is either
or
modulo
. However, we seek the number of
, and
. By CRT,
is either
or
modulo
, and the answer is
.
-TheUltimate123
Step Solution
Clearly is an integer. The part we need to verify as an integer is, upon common denominator,
. Clearly, the numerator must be even for the fraction to be an integer. Therefore,
is even and n is even, aka
for some integer
. Then, we can substitute
and see that
is trivially integral. Then, substitute for the rest of the non-confirmed-integral terms and get
. 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
, and we see that
for some integer
. From there we now know that
. 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
, so combining with divisibility by 6,
is
or
(mod
). There are
cases for each, hence the answer
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.