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I've just started hanging out with my friend Agatha, but we don't know what do to. Suddenly, she comes up with an idea: A game of numbers. These are the rules:

  • Pick a number between 7 and 100, inclusive.
  • Take the prime factorization of that number, in exponential form (i.e. express it as $p_1^{e_1}p_2^{e_2}...$). Take all the $p_i$ and $e_i$, and choose either their sum or their product: your number becomes the chosen value.
  • Keep doing this until either your number becomes less than 7, you end up with a number that you have already chosen earlier this turn, or go over your limit.
  • You lose if your opponent lasts longer than you, and your limit (on starting numbers) is increased by 10. In case of a tie, simply repeat the round.

What are the optimal numbers if you're going to play 10 rounds?

I don't know the answer to this.

BONUS: What is the smallest possible starting number with a chain length of at least $n$, up to $n = 10$?

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  • 1
    $\begingroup$ Is this an original puzzle? If not, we require attribution, such as a note of where you go the puzzle or a link to the original source. Also, what does "Take all... another unique number" mean? $\endgroup$ – bobble Jun 20 at 15:52
  • $\begingroup$ Presumably, it refers to all the parts of the expansion. I'm curious about the fourth bullet point, is the goal to have as few remaining moves as possible? $\endgroup$ – AxiomaticSystem Jun 20 at 19:34
  • $\begingroup$ @bobble: I'm sorry, I don't know if any exist. That part means to use all the numbers shown (i.e. for 24 it would be 2, 3, and 3.) $\endgroup$ – Player1456 Jun 21 at 2:15
  • $\begingroup$ @AxiomaticSystem: Quite the opposite: you lose if your opponent lasts longer before ending their turn. $\endgroup$ – Player1456 Jun 21 at 2:15
  • $\begingroup$ What is the purpose of increasing the limit when losing? $\endgroup$ – justhalf Jun 21 at 2:19
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First, note that

Your number will never increase:
If you have $n = p_1^{e_1}p_2^{e_2}...$, then $\sum\limits_i p_i+e_i \leq \prod\limits_i p_ie_i \leq \prod\limits_i p_i^{e_i}$.
(The former equality happens for primes and $4$, the latter happens for squarefree numbers and twice squarefree numbers.)
Then a simple computer search confirms that you can never make more than six moves from any number at most $200$ (which would be your limit if you lost every round.) The only such chain starting below $100$ begins at $72$: $72 \rightarrow 36 \rightarrow 24 \rightarrow 18 \rightarrow 12 \rightarrow 7 \rightarrow 7$
Assuming you can't simply pick $72$ every round, the other numbers admitting chains of length six are $108, 144, 152, 155, 171, 180, 186,$ and $192$.

As for a strategy, it depends on the circumstances:

If numbers can be reused, the game is clearly a draw.
Otherwise, each player essentially has a list of usable numbers, sorted by the length of their corresponding chain. Losing a round adds more numbers to the list, and the winner of a round is essentially determined by who has more six- (or five-) move numbers in their list. [TODO: Specifics.]

Bonus Time!

$n = 2,...,16: 7, 10, 18, 24, 36, 72, 248, 496, 1044, 2088, 7272, 16624, 33328, 74916, 149832.$
Specifics: $7 = 7$
$10 = 2 × 5 \rightarrow 2 + 5 = 7$
$18 = 2 × 3^2 \rightarrow 2 × 3 × 2 = 2^2 \times 3 \rightarrow 2 + 2 + 3 = 7$
$24 = 2^3 × 3 \rightarrow 2 × 3 × 3 = 18$
$36 = 2^2 × 3^2 \rightarrow 2 × 2 × 3 × 2 = 24$
$72 = 2^3 × 3^2 \rightarrow 2 × 3 × 3 × 2 = 36$
$248 = 2^3 × 31 \rightarrow 2 × 3 × 31 \rightarrow 2 + 3 + 31 = 36$
$496 = 2^4 × 31\rightarrow 2 × 4 × 31 = 248$
$1044 = 2^2 × 3^2 × 29 \rightarrow 2 × 2 × 3 × 2 × 29 \rightarrow 2 × 3 × 3 × 29 \rightarrow 2 × 3 × 2 × 29 \rightarrow 2 + 2 + 3 + 29 = 36$
$2088 = 2^3 × 3^2 × 29 \rightarrow 2 × 3 × 3 × 2 × 29 = 1044$
$7272 = 2^3 × 3^2 × 101 \rightarrow 2 × 3 × 3 × 2 × 101 \rightarrow 2×2×3×2×101 \rightarrow 2×3×3×101 \rightarrow 2×3×2×101 \rightarrow 2+3+2+101 = 108 = 2^2×3^3 \rightarrow 2×2×3×3 = 36$
$16624 = 2^4 × 1039 \rightarrow 2×4×1039 \rightarrow 2×3×1039 \rightarrow 2+3+1039 = 1044$
$33328 = 2^4×2083 \rightarrow 2×4×2083 \rightarrow 2×3×2083 \rightarrow 2+3+2083 = 2088$
$74916 = 2^2×3^2×2081$... You get the idea.

| improve this answer | |
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  • $\begingroup$ You've figured it all out! Congratulations. $\endgroup$ – Player1456 Jun 23 at 4:25
  • $\begingroup$ OP's comment to the puzzle implies that 1's in exponents are ignored, otherwise increases would be possible. $\endgroup$ – AxiomaticSystem 2 days ago
  • $\begingroup$ @AxiomaticSystem Ah, I didn't see that comment, apologies. $\endgroup$ – hexomino 2 days ago

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