Difference between revisions of "2016 AMC 12B Problems/Problem 25"
(→Solution 3: Remove. 1, 2, 5, 21, ... is not Catalan, and the rest of the argument is wrong as well. Two wrongs make a right, I guess) |
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==Solution 2== | ==Solution 2== | ||
− | Since the product <math>a_1a_2\cdots a_k</math> is an integer, | + | 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>. |
− | |||
− | |||
− | <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,-2,-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>. | ||
+ | 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:</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. | ||
+ | |||
+ | ==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}} |
Revision as of 21:54, 6 November 2021
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: 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 |
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