2018 AMC 10B Problems/Problem 20
- The following problem is from both the 2018 AMC 12B #18 and 2018 AMC 10B #20, so both problems redirect to this page.
Contents
Problem
A function is defined recursively by
and
for all integers
. What is
?
Solution 1 (Algebra)
For all integers note that
It follows that
~MRENTHUSIASM
Solution 2 (Algebra)
For all integers we rearrange the given equation:
For all integers
it follows that
For all integers
we add
and
For all integers
it follows that
For all integers
we subtract
from
From
we have the following system of
equations:
We add these equations up to get
from which
~AopsUser101 ~MRENTHUSIASM
Solution 3 (Finite Differences)
Preamble: In this solution, we define the sequence to satisfy
where
represents the
th term of the sequence
This solution will show a few different perspectives. Even though it may not be as quick as some of the solutions above, I feel like it is an interesting concept, and may be more motivated.
To begin, we consider the sequence formed when we take the difference of consecutive terms between
Define
Notice that for
we have
Notice that subtracting the second equation from the first, we see that
If you didn’t notice that repeated directly in the solution above, you could also, possibly more naturally, take the finite differences of the sequence
so that you could define
Using a similar method as above through reindexing and then subtracting, you could find that
The sum of any six consecutive terms of a sequence which satisfies such a recursion is
in which you have that
In the case in which finite differences didn’t reduce to such a special recursion, you could still find the first few terms of
to see if there are any patterns, now that you have a much simpler sequence. Doing so in this case, it can also be seen by seeing that the sequence
looks like
in which the same result follows.
Using the fact that repeats every six terms, this motivates us to look at the sequence
more carefully. Doing so, we see that
looks like
(If you tried pattern finding on sequence
directly, you could also arrive at this result, although I figured defining a second sequence based on finite differences was more motivated.)
Now, there are two ways to finish.
Finish Method #1: Notice that any six consecutive terms of sum to
after which we see that
Therefore,
Finish Method #2: Notice that
~Professor-Mom
Solution 4 (Bash)
Start out by listing some terms of the sequence.
Notice that
whenever
is an odd multiple of
, and the pattern of numbers that follow will always be
,
,
,
,
,
.
The largest odd multiple of
smaller than
is
, so we have
minor edits by bunny1
Solution 5 (Bash)
Writing out the first few values, we get:
. Examining, we see that every number
where
has
,
, and
. The greatest number that's
and less
is
, so we have
.
Video Solution
https://www.youtube.com/watch?v=aubDsjVFFTc
~bunny1
~savannahsolver
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 10 Problems and Solutions |
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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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