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== | + | ==Solution 1 (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>. | ||
Revision as of 22:41, 21 February 2018
Problem
Find the number of positive integers less than
such that
is an integer.
Solution 1 (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 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.