Xmasium

Dec 2019
52
1
ok
In one of Santa's research laboratories, the scientists have discovered a new chemical element and have named it Xmasium (by analogy with the famous elements Rubidium, Caesium, and Francium). Xmasium exists in three different types: There is α-Xmasium, β-Xmasium, and γ-Xmasium. If two Xmasium-atoms of different type collide, they sometimes merge into a single atom of the third type. If two Xmasium-atoms of the same type collide, they repel each other and nothing else happens.

Ruprecht puts 91 α-Xmasium-atoms, 25 β-Xmasium-atoms, and 4 γ-Xmasium-atoms into a cooking pot, covers it, and then leaves for lunch. When he returns, he notices that all Xmasium-atoms in the pot now are of the same type.

What is the largest possible number z of Xmasium-atoms in the pot at that moment?

  1. This largest possible number satisfies 20 ≤ z≤ 23.

  2. This largest possible number satisfies 24 ≤ z≤ 27.

  3. This largest possible number satisfies 28 ≤ z≤ 31.

  4. This largest possible number satisfies 32 ≤ z≤ 35.

  5. This largest possible number satisfies 36 ≤ z≤ 39.

  6. This largest possible number satisfies 40 ≤ z≤ 43.

  7. This largest possible number satisfies 44 ≤ z≤ 47.

  8. This largest possible number satisfies 48 ≤ z≤ 51.

  9. This largest possible number satisfies 52 ≤ z≤ 55.

  10. This largest possible number satisfies 56 ≤ z≤ 59.
 
Jun 2019
493
260
USA
Ignoring how this can possibly satisfy conservation of mass, I assume the largest number of atoms remaining would be if they were all $\alpha$. You'd need an equal amount of $\beta$ and $\gamma$ before the last step, so...
$10\alpha + 10\beta \rightarrow 10 \gamma \Rightarrow 81\alpha + 15\beta + 14\gamma$
$14\beta+14\gamma\rightarrow 14\alpha \Rightarrow 95\alpha +1\beta$

If I collide an $\alpha$ with the $\beta$, I'll have a leftover $\gamma$. If I collide this with another $\alpha$, I'll have a leftover $\beta$. No way to make this work out, so either I've missed something, or my original assumption was faulty.

Let's try for all $\beta$.
$l\alpha +l\beta\rightarrow l\gamma \Rightarrow (95-l)\alpha +(25-l)\beta +(4+l)\gamma$
No matter what I choose for $l$, one of ${\alpha,\gamma}$ will be even and the other odd, so they can't be equal. I can't fix this with extra $\beta$s, either.

What about all $\gamma$?
$25\alpha+25\beta\rightarrow 25\gamma \Rightarrow 66\alpha +29\gamma$
$29\alpha+29\gamma\rightarrow 29\beta \Rightarrow 37\alpha+29\beta$
$29\alpha+29\beta\rightarrow 29\gamma \Rightarrow 8\alpha+29\gamma$
$4\alpha+4\gamma\rightarrow 4\beta \Rightarrow 4\alpha+4\beta+25\gamma$
$4\alpha+4\beta\rightarrow 4\gamma \Rightarrow 29\gamma$

That would be my initial guess, but I have a presentation to attend in 5 minutes, so I don't have time to check my work or formalise it to prove there wasn't a more efficient path.