2017 AIME II Problems/Problem 8
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 |
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