2016 AMC 12B Problems/Problem 25
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) | |
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