Twenty-One
Virgil should go second. Whatever Scott adds, Virgil should then add to make four. For example, if Scott says 3, Virgil should say 1.
Using this strategy, Virgil will say 4, 8, 12, 16 then 20, forcing Scott to go above 21.
Extension
(i) If instead of 21, 22 cannot be said/beaten, how should Virgil win? How about 23? Or 24? How about \(n\)?
(ii) If instead of adding 1 to 3, 1 to 4 can be added, how should Virgil win? How about 1 to 5? Or 2 to 5? How about \(m\) to \(l\)?
(iii) Alan wants to join the game. Can Virgil win if there are three people? Can he win if there are \(k\) people?
Odd and Even Outputs
| \(n\) |
odd | even |
\(m\) | odd | odd | odd |
e | odd | odd |
\(g(n,m)=1\)
| \(n\) |
odd | even |
\(m\) | odd | odd | odd |
e | odd | even |
\(g(n,m)=n\times m + n + m\)
| \(n\) |
odd | even |
\(m\) | odd | odd | odd |
e | even | odd |
\(g(n,m)=n\times m +n+1\)
| \(n\) |
odd | even |
\(m\) | odd | odd | odd |
e | even | even |
\(g(n,m)=m\)
| \(n\) |
odd | even |
\(m\) | odd | odd | even |
e | odd | odd |
\(g(n,m)=n\times m+m+1\)
| \(n\) |
odd | even |
\(m\) | odd | odd | even |
e | odd | even |
\(g(n,m)=n\)
| \(n\) |
odd | even |
\(m\) | odd | odd | even |
e | even | odd |
\(g(n,m)=n+m+1\)
| \(n\) |
odd | even |
\(m\) | odd | odd | even |
e | even | even |
\(g(n,m)=n\times m\)
|
| \(n\) |
odd | even |
\(m\) | odd | even | odd |
e | odd | odd |
\(g(n,m)=n\times m+1\)
| \(n\) |
odd | even |
\(m\) | odd | even | odd |
e | odd | even |
\(g(n,m)=n+m\)
| \(n\) |
odd | even |
\(m\) | odd | even | odd |
e | even | odd |
\(g(n,m)=n+1\)
| \(n\) |
odd | even |
\(m\) | odd | even | odd |
e | even | even |
\(g(n,m)=n\times m+n\)
|
| \(n\) |
odd | even |
\(m\) | odd | even | even |
e | odd | odd |
\(g(n,m)=m+1\)
| \(n\) |
odd | even |
\(m\) | odd | even | even |
e | odd | even |
\(g(n,m)=n\times m+n\)
| \(n\) |
odd | even |
\(m\) | odd | even | even |
e | even | odd |
\(g(n,m)=n\times m+n+m+1\)
| \(n\) |
odd | even |
\(m\) | odd | even | even |
e | even | even |
\(g(n,m)=2\)
Extension
Can you find functions \(h:\mathbb{N}\times\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}\) (call the inputs \(n\), \(m\) and \(l\)) to give the following outputs:
\(l\) odd
| \(n\) |
odd | even |
\(m\) | odd | even | even |
e | even | even |
|
\(l\) even
| \(n\) |
odd | even |
\(m\) | odd | even | even |
e | even | even |
|
\(l\) odd
| \(n\) |
odd | even |
\(m\) | odd | even | even |
e | even | even |
|
\(l\) even
| \(n\) |
odd | even |
\(m\) | odd | even | even |
e | even | odd |
|
etc
|
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