Difference between revisions of "2020 AMC 8 Problems/Problem 22"
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==Video Solutions== | ==Video Solutions== | ||
+ | https://youtu.be/SPNobOd4t1c (Channel also has resources to prepare for your AIME qualification) | ||
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https://youtu.be/lhDFmiKNPBg | https://youtu.be/lhDFmiKNPBg | ||
Revision as of 14:24, 3 January 2021
Contents
Problem 22
When a positive integer is fed into a machine, the output is a number calculated according to the rule shown below.
For example, starting with an input of
the machine will output
Then if the output is repeatedly inserted into the machine five more times, the final output is
When the same
-step process is applied to a different starting value of
the final output is
What is the sum of all such integers
Solution 1
We start with the final output of and work backwards, taking care to consider all possible inputs that could have resulted in any particular output. This produces the following set of possibilities at each stage:
where, for example,
must come from
(as there is no integer
satisfying
), but
could come from
or
(as
, and
is even while
is odd). By construction, the last set in this sequence contains all the numbers which will lead to the number
at the end of the
-step process, and their sum is
.
Solution 2 (variant of Solution 1)
As in Solution 1, we work backwards from , this time showing the possible cases in a tree diagram:
The possible numbers are those at the "leaves" of the tree (the ends of the various branches), which are ,
,
, and
. Thus the answer is
.
Solution 3 (algebraic)
We begin by finding the inverse of the function that the machine uses. Call the input and the output
. If
is even,
, and if
is odd,
. We can therefore see that
when
is even and
when
is odd. Therefore, starting with
, if
is even,
, and if
is odd,
, but the latter is not valid since
is not actually odd. This means that the 2nd-to-last number in the sequence has to be
. Now, substituting
into the inverse formulae, if
is even,
(which is indeed even), and if
is odd,
, which is not an integer. This means the 3rd-to-last number in the sequence has to be
. Substituting in
, if
is even,
, but if
is odd,
. Both of these are valid solutions, so the 4th-to-last number can be either
or
. If it is
, then by the argument we have just made, the 5th-to-last number has to be
, the 6th-to-last number has to be
, and the 7th-to-last number, which is the first number, must be either
or
. In this way, we have ultimately found two solutions:
and
.
On the other hand, if the 4th-to-last number is , substituting
into the inverse formulae shows that the 5th-to-last number is either
or
, but the latter is not an integer. Substituting
shows that if
is even,
, but if I is odd,
, and both of these are valid. If the 6th-to-last number is
, then the first number must be
, since
is not an integer; if the 6th-to-last number is
then the first number has to be
, as
is not an integer. This means that, in total, there are
solutions for
, specifically,
,
,
, and
, which sum to
.
Video Solutions
https://youtu.be/SPNobOd4t1c (Channel also has resources to prepare for your AIME qualification)
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 AJHSME/AMC 8 Problems and Solutions |
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