Difference between revisions of "2017 AIME II Problems/Problem 8"
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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 \equiv 0 \pmod{5}</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 if <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>. | 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 \equiv 0 \pmod{5}</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 if <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 \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 | + | 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>. |
=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 18:32, 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 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
.
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.