+1206 We define a function f(x) such that f(11)=34, and if there exists an integer a such that f(a)=b, then f(b) is defined
and f(b)=3b+1 if b is odd
\(f(b)=\frac{b}{2}\) if b is even.
What is the smallest possible number of integers in the domain of f?
+9466 We are given that f(11) = 34 so we know that 11 is in the domain.
domain so far = {11}
There exists an integer 11 such that f(11) = 34 , so f(34) is defined. And so 34 is in the domain.
domain so far = {11, 34}
So notice that just like we had to add the result of f(11) to the domain, we will now have to add the result of f(34) to the domain (then the result of that, and the result of that, etc.).
| f(34) = 17 | _____ | domain so far = {11, 34, 17} |
| f(17) = 52 |
| domain so far = {11, 34, 17, 52} |
| f(52) = 26 | domain so far = {11, 34, 17, 52, 26} | |
| f(26) = 13 |
| domain so far = {11, 34, 17, 52, 26, 13} |
| f(13) = 40 | domain so far = {11, 34, 17, 52, 26, 13, 40} | |
| f(40) = 20 |
| domain so far = {11, 34, 17, 52, 26, 13, 40, 20} |
| f(20) = 10 | domain so far = {11, 34, 17, 52, 26, 13, 40, 20, 10} | |
| f(10) = 5 |
| domain so far = {11, 34, 17, 52, 26, 13, 40, 20, 10, 5} |
| f(5) = 16 | domain so far = {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16} | |
| f(16) = 8 |
| domain so far = {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8} |
| f(8) = 4 | domain so far = {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4} | |
| f(4) = 2 |
| domain so far = {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2} |
| f(2) = 1 | domain so far = {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1} | |
| f(1) = 4 |
| domain so far = {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1} |
Finally, we have reached a point where we do not have to add anything else to the domain.
4 is already in the set, so we do not have to add anything else.
So the smallest possible number of integers in the domain is 15
(Thank you CPhill for correcting my previous mistake!)
+9466 We are given that f(11) = 34 so we know that 11 is in the domain.
domain so far = {11}
There exists an integer 11 such that f(11) = 34 , so f(34) is defined. And so 34 is in the domain.
domain so far = {11, 34}
So notice that just like we had to add the result of f(11) to the domain, we will now have to add the result of f(34) to the domain (then the result of that, and the result of that, etc.).
| f(34) = 17 | _____ | domain so far = {11, 34, 17} |
| f(17) = 52 |
| domain so far = {11, 34, 17, 52} |
| f(52) = 26 | domain so far = {11, 34, 17, 52, 26} | |
| f(26) = 13 |
| domain so far = {11, 34, 17, 52, 26, 13} |
| f(13) = 40 | domain so far = {11, 34, 17, 52, 26, 13, 40} | |
| f(40) = 20 |
| domain so far = {11, 34, 17, 52, 26, 13, 40, 20} |
| f(20) = 10 | domain so far = {11, 34, 17, 52, 26, 13, 40, 20, 10} | |
| f(10) = 5 |
| domain so far = {11, 34, 17, 52, 26, 13, 40, 20, 10, 5} |
| f(5) = 16 | domain so far = {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16} | |
| f(16) = 8 |
| domain so far = {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8} |
| f(8) = 4 | domain so far = {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4} | |
| f(4) = 2 |
| domain so far = {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2} |
| f(2) = 1 | domain so far = {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1} | |
| f(1) = 4 |
| domain so far = {11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1} |
Finally, we have reached a point where we do not have to add anything else to the domain.
4 is already in the set, so we do not have to add anything else.
So the smallest possible number of integers in the domain is 15
(Thank you CPhill for correcting my previous mistake!)
