Difference between revisions of "2016 AMC 12B Problems/Problem 25"
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<cmath>,1,,5,,2,,9,,-1,,-3,,8,,-5,,0,\dots</cmath> | <cmath>,1,,5,,2,,9,,-1,,-3,,8,,-5,,0,\dots</cmath> | ||
<math>c_n</math> is first a multiple of <math>19</math> at <math>n = \boxed{\textbf{(A)}\ 17}</math>. ~[[User:emerald_block|emerald_block]] | <math>c_n</math> is first a multiple of <math>19</math> at <math>n = \boxed{\textbf{(A)}\ 17}</math>. ~[[User:emerald_block|emerald_block]] | ||
+ | |||
+ | ==Solution 4 (Using a formula) | ||
+ | |||
+ | Consider the product <math>a_1a_2\cdots a_k</math> (will finish tommorow) | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|after=Last Problem|num-b=24}} | {{AMC12 box|year=2016|ab=B|after=Last Problem|num-b=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:57, 12 September 2024
Contents
[hide]Problem
The sequence is defined recursively by
,
, and
for
. What is the smallest positive integer
such that the product
is an integer?
Solution 1
Let . Then
and
for all
. The characteristic polynomial of this linear recurrence is
, which has roots
and
.
Therefore, for constants to be determined
. Using the fact that
we can solve a pair of linear equations for
:
.
Thus ,
, and
.
Now, , so we are looking for the least value of
so that
.
Note that we can multiply all by three for convenience, as the
are always integers, and it does not affect divisibility by
.
Now, for all even the sum (adjusted by a factor of three) is
. The smallest
for which this is a multiple of
is
by Fermat's Little Theorem, as it is seen with further testing that
is a primitive root
.
Now, assume is odd. Then the sum (again adjusted by a factor of three) is
. The smallest
for which this is a multiple of
is
, by the same reasons. Thus, the minimal value of
is
.
Solution 2
Since the product is an integer, it must be a power of
, so the sum of the base-
logarithms must be an integer. Multiply all of these logarithms by
(to make them integers), so the sum must be a multiple of
.
The logarithms are . Using the recursion
(modulo
to save calculation time), we get the sequence
Listing the numbers out is expedited if you notice
.
The cycle repeats every terms. Notice that since
, the first
terms sum up to a multiple of
. Since
, we only need at most the first
terms to sum up to a multiple of
, and this is the lowest answer choice.
Note 1: To rigorously prove this is the smallest value, you will have to keep a running sum of the terms and check that it is never a multiple of before the
th term.
Note 2: In response to note 1, it can be proven that , where
. Since
is a multiple of
, it suffices to find the minimal
such that
. In this case,
happens to be minimal such
, so the answer would be
.
The relation can be proven by rearranging the relation
to
for all integers
, then adding those
equations together. The LHS telescopes into
, and the RHS becomes
. Therefore, if you don't find a cleaner solution involving the relation
, you can always solve the problem just by considering the value of
rather than keeping a running sum.
Solution 3
Like in Solution 2, calculate the first few terms of the sequence, but also keep a running sum of the logarithms (not modulo
here):
Notice that
for odd
and
for even
. Since
is relatively prime to
, we can ignore even
and calculate odd
using
(modulo
):
is first a multiple of
at
. ~emerald_block
==Solution 4 (Using a formula)
Consider the product (will finish tommorow)
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.