# The Game of Numbers (#01)

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 $$n = 10$$?

• 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? – bobble Jun 20 at 15:52
• 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? – AxiomaticSystem Jun 20 at 19:34
• @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.) – Player1456 Jun 21 at 2:15
• @AxiomaticSystem: Quite the opposite: you lose if your opponent lasts longer before ending their turn. – Player1456 Jun 21 at 2:15
• What is the purpose of increasing the limit when losing? – justhalf Jun 21 at 2:19

First, note that

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.

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