Difference between revisions of "2016 AMC 12B Problems/Problem 25"
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The logarithms are <math>b_n = 19\log_2 a_n</math>. Using the recursion <math>b_0 = 0, b_1 = 1, b_n = b_{n-1}+2b_{n-2}</math> (modulo <math>19</math> to save calculation time), we get the sequence | The logarithms are <math>b_n = 19\log_2 a_n</math>. Using the recursion <math>b_0 = 0, b_1 = 1, b_n = b_{n-1}+2b_{n-2}</math> (modulo <math>19</math> to save calculation time), we get the sequence | ||
− | <cmath>0,1,1,3,5,11,2,5,9,0,-1,-1,-2,-5,-11,- | + | <cmath>0,1,1,3,5,11,2,5,9,0,-1,-1,-2,-5,-11,-3,-5,-9,0,\dots</cmath> |
Listing the numbers out is expedited if you notice <math>b_{n+1}=2b_n+(-1)^n</math>. | Listing the numbers out is expedited if you notice <math>b_{n+1}=2b_n+(-1)^n</math>. | ||
Revision as of 23:44, 31 August 2023
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: 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.
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
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.