2020 AMC 8 Problems/Problem 22
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 (Includes all the problems and has a free class update)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.