Difference between revisions of "2016 AMC 12B Problems/Problem 25"
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Now, for all even <math>k</math> the sum (adjusted by a factor of three) is <math>2^1+2^2+\cdots+2^k=2^{k+1}-2</math>. The smallest <math>k</math> for which this is a multiple of <math>19</math> is <math>k=18</math> by Fermat's Little Theorem, as it is seen with further testing that <math>2</math> is a primitive root <math>\pmod{19}</math>. | Now, for all even <math>k</math> the sum (adjusted by a factor of three) is <math>2^1+2^2+\cdots+2^k=2^{k+1}-2</math>. The smallest <math>k</math> for which this is a multiple of <math>19</math> is <math>k=18</math> by Fermat's Little Theorem, as it is seen with further testing that <math>2</math> is a primitive root <math>\pmod{19}</math>. | ||
− | Now, assume <math>k</math> is odd. Then the sum (again adjusted by a factor of three) is <math>2^1+2^2+\cdots+2^k+1=2^{k+1}-1</math>. The smallest <math>k</math> for which this is a multiple of <math>19</math> is <math>k=17</math>, by the same reasons. Thus, the minimal value of <math>k</math> is <math>\textbf{(A) } 17</math>. | + | Now, assume <math>k</math> is odd. Then the sum (again adjusted by a factor of three) is <math>2^1+2^2+\cdots+2^k+1=2^{k+1}-1</math>. The smallest <math>k</math> for which this is a multiple of <math>19</math> is <math>k=17</math>, by the same reasons. Thus, the minimal value of <math>k</math> is <math>\boxed{\textbf{(A) } 17}</math>. |
==Solution 2== | ==Solution 2== | ||
− | Since the product <math>a_1a_2\cdots a_k</math> is an integer, the sum of the | + | Since the product <math>a_1a_2\cdots a_k</math> is an integer, it must be a power of <math>2</math>, so the sum of the base-<math>2</math> logarithms must be an integer. Multiply all of these logarithms by <math>19</math> (to make them integers), so the sum must be a multiple of <math>19</math>. |
− | |||
− | |||
+ | 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,-3,-5,-11,-2,-5,-9,0,\dots</cmath> | ||
+ | Listing the numbers out is expedited if you notice <math>b_{n+1}=2b_n+(-1)^n</math>. | ||
− | <math> | + | The cycle repeats every <math>9+9=18</math> terms. Notice that since <math>b_n+b_{n+9} \equiv 0 \pmod{19}</math>, the first <math>18</math> terms sum up to a multiple of <math>19</math>. Since <math>b_{18}=0</math>, we only need at most the first <math>\boxed{\textbf{(A)}\ 17}</math> terms to sum up to a multiple of <math>19</math>, and this is the lowest answer choice. |
+ | |||
+ | <b>Note 1:</b> 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 <math>19</math> before the <math>17</math>th term. | ||
+ | |||
+ | <b>Note 2:</b> In response to note 1, it can be proven that <math>b_{n+2} = 2S + 1</math>, where <math>S = \sum^{n}_{i=1} b_i</math>. Since <math>S</math> is a multiple of <math>19</math>, it suffices to find the minimal <math>n \geq 1</math> such that <math>b_{n+2} \equiv 1 \pmod {19} </math>. In this case, <math>n = 17</math> happens to be minimal such <math>n</math>, so the answer would be <math>17</math>. | ||
+ | |||
+ | The relation <math>b_{n+2} = 2S + 1</math> can be proven by rearranging the relation <math>b_{i+2} = b_{i+1} + 2b_i</math> to <math>b_{i+2} - b_{i+1} = 2b_i</math> for all integers <math>0 \leq i \leq n</math>, then adding those <math>n+1</math> equations together. The LHS telescopes into <math>b_{n+2} - 1</math>, and the RHS becomes <math>2S</math>. Therefore, if you don't find a cleaner solution involving the relation <math>b_n+b_{n+9} \equiv 0 \pmod{19}</math>, you can always solve the problem just by considering the value of <math>b_{n+2}</math> rather than keeping a running sum. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Like in [[#Solution 2|Solution 2]], calculate the first few terms of the sequence, but also keep a running sum <math>c_n</math> of the logarithms (not modulo <math>19</math> here): | ||
+ | <cmath>0,1,2,5,10,21,42,\dots</cmath> | ||
+ | Notice that <math>c_n=2c_{n-1}+1</math> for odd <math>n</math> and <math>c_n=2c_{n-1}</math> for even <math>n</math>. Since <math>2</math> is relatively prime to <math>19</math>, we can ignore even <math>n</math> and calculate odd <math>n</math> using <math>c_1 = 1, c_{n} = 4c_{n-2}+1</math> (modulo <math>19</math>): | ||
+ | <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]] | ||
==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}} |
Latest revision as of 15:07, 6 July 2024
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
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 |
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