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== | + | ==Solution 1== |
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> | 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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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\equiv0(\text{mod }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 the equation <math>n^6+6n^5+30n^4+120n^3+360n^2=720m</math> mod <math>3</math>, the result is <math>n^6\equiv0(\text{mod }3)</math>. Therefore <math>n</math> must be a multiple of <math>3</math>. If <math>n</math> is a multiply of three, that means <math>n</math> can be written in the form <math>3b</math> where <math>b</math> is a positive integer. Replacing <math>n</math> with <math>3b</math> in the expression, <math>729b^6+1458b^5+2430b^4+3240b^3+3240b^2</math> is divisible by <math>9</math> because each coefficient is divisible by <math>9</math>. Therefore, if <math>n</math> is a multiple of <math>3</math>, <math>n^6+6n^5+30n^4+120n^3+360n^2</math> is divisibly by <math>9</math>. | ||
+ | Taking the equation <math>n^6+6n^5+30n^4+120n^3+360n^2=720m</math> mod <math>5</math>, the result is <math>n^6+n^5\equiv0(\text{mod }3)</math>. The only values of <math>n (\text{mod }5)</math> that satisfy the equation are <math>n\equiv0(\text{mod }5)</math> and <math>n\equiv4(\text{mod }5)</math>. Therefore is <math>n</math> is <math>0</math> or <math>4</math> mod <math>5</math>, <math>n^6+6n^5+30n^4+120n^3+360n^2</math> will be a multiple of <math>5</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\times9\times5</math> is to have <math>n\equiv0(\text{mod }2)</math>, <math>n\equiv0(\text{mod }3)</math>, and <math>n\equiv0,4(\text{mod }5)</math>. By the Chinese Remainder Theorem or simple guessing and checking, we see <math>n\equiv0,24(\text{mod }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>. | ||
=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 17:37, 23 March 2017
Contents
Problem
Find the number of positive integers less than
such that
is an integer.
Solution 1
The denominator contains . Therefore, one possibility is that
. This yields the numbers
. There are a total of
numbers in the sequence. We express the last two terms as
This yields that
. Therefore, we get the final answer of
Solution 2
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 multiply 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 is
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
.
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.