Game Spinner Maximum Spins
John and Gary are playing a game. John spins a spinner numbered with integers from 1 to 20. Gary then writes a list of all of the positive factors of the number spun except for the number itself. Gary then creates a new spinner with all of the numbers on his list. John then spins this spinner, and the process continues. The game is over when the spinner has no numbers on it. If John spins a 20 on his first spin, what is the maximum number of total spins (including the one he already made) that John can make before the game is over?
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
If John spins a 20, then Gary's list contains the numbers 1, 2, 4, 5, 10. Thus, these are the numbers on the second spinner.
If John spins a 1, then Gary's list will be empty because there are no positive factors of 1 besides itself. Thus, the game will be over. This yields a maximum of 1 additional spin.
If John spins a 2, then Gary's list will only contain the number 1. Then on John's next spin, we will have the same scenario as above. This yields a maximum of 2 additional spins.
If John spins a 4, then Gary's list will contain the numbers 1 and 2. As we have already found above, spinning a 2 yields more additional spins than a 1, so the maximum additional spins in this case is 3 spins.
If John spins a 5, then Gary's list will only contain the number 1. As above, this will yield a maximum of 2 additional spins.
Finally, if John spins a 10, then Gary's list will contain the numbers 1, 2, and 5. Of these numbers, 2 and 5 have the highest maximum number of additional spins, so this case has a maximum of 3 additional spins.
Thus, of all of the possibilities, spinning a 4 or 10 next could result in 3 additional spins, so the maximum total number of spins is $ \boxed{4} $. These would be achieved by spinning 20, 10, 2, 1 or 20, 10, 5, 1 or 20, 4, 2, 1.
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