# 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? Jun 20 '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? Jun 20 '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.) Jun 21 '20 at 2:15
• @AxiomaticSystem: Quite the opposite: you lose if your opponent lasts longer before ending their turn. Jun 21 '20 at 2:15
• What is the purpose of increasing the limit when losing? Jun 21 '20 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. Jun 23 '20 at 4:25
• OP's comment to the puzzle implies that 1's in exponents are ignored, otherwise increases would be possible. Sep 24 '20 at 11:37
• @AxiomaticSystem Ah, I didn't see that comment, apologies. Sep 24 '20 at 12:30