Dice Dicey Dice Dice

I posit that it is possible if and only if $N$ only has prime factors $2$ , $3$ and $5$ .

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Firstly, let's prove that if $N$ has prime factors that are not $2$ , $3$ or $5$ , then it is impossible. This is rather easy to prove. For an equal probability of totalling an integer $j$ between $1$ to $N$ given a set of numbered dice, the number of rolls that result in $j$ is the same regardless of $j$ . Hence the number of possible rolls $A$ is divisible by $N$ .

$A$ is also equivalent to the product of the number of sides of the dice used. For instance, if an 8-die and a 6-die is used, $A=48$ . Since the allowed number of sides for the dice only has prime factors $2,3,5$ , $A$ would similarly only have prime factors $2,3,5$ .

Since $N$ is a divisor of $A$ , $N$ would also only have prime factors $2,3,5$ . Hence proved.

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Now we need to prove that for every $N$ with prime factors $2, 3, 5$ , there is a dice numbering that satisfies Vi's requirements. For this we simply construct a solution:

For $N=5^A \cdot 3^B \cdot 2^C$ , the dice numbering can be:

• $A$ 20-sided dice numbered $\left\{ 0, 5^n, 2 \cdot 5^n, 3 \cdot 5^n, 4 \cdot 5^n \right\}$ , each element repeated 4 times to fill up all 20 sides, for $0 \le n \le A-1$
• $B$ 6-sided dice numbered $\left\{ 0, 3^n 5^A, 2 \cdot 3^n 5^A \right\}$ , each element repeated 2 times to fill up all 6 sides, $0 \le n \le B-1$
• $C-1$ 4-sided dice numbered $\left\{ 0, 2^n 3^B 5^A \right\}$ , each element repeated 2 times to fill up all 4 sides, $1 \le n \le C-1$
• One 4-sided die numbered $\left\{ 1, 3^B 5^A + 1 \right\}$ , each element repeated 2 times to fill up all 4 sides

Note that this requires $A+B+C$ dice, which isn't the most optimal construction. But it works.

To show why this works, we can connect this problem to Number Bases .

Firstly, it would be prudent to modify the problem from "pick an integer from $1$ to $N$ with equal probability" to "pick an integer from $0$ to $N-1$ with equal probability". This would simplify the problem as would be seen in the later steps. The approach would be to use $N$ 's representation in certain number bases to find a dice numbering.

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It would be simpler to understand the approach (and easier for me to express with clarity) if we considered a simple case of $N=5^2=25$ . We want to find a dice numbering that would pick an integer from $0$ to $24$ with equal probability. We can do so by representing $24$ in base 5: $24_{10} = 44_{5}$ . Now consider a number $\bar{ab}_5$ . We can consider every possible combination of $a$ and $b$ to be any integer between $0$ and $4$ inclusive, and this would generate an integer from $0$ to $44_5$ exactly once. Converting that to a dice numbering we have

• $2$ 20-sided dice numbered $\left\{ 0, 5^n, 2 \cdot 5^n, 3 \cdot 5^n, 4 \cdot 5^n \right\}$ , each element repeated 4 times to fill up all 20 sides, for $0 \le n \le 1$

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Now to generalise to $5^A \cdot 3^B \cdot 2^C$ , we can modify the argument to consider the expansion as sort of a pseudo-number-base-representation-thing:

$5^A \cdot 3^B \cdot 2^C - 1 = 4 + 4 \cdot 5 + 4 \cdot 5^2 ... + 4 \cdot 5^{A-1} + 5^A\left[ 2 + 2 \cdot 3 + 2 \cdot 3^2 ... + 2 \cdot 3^{B-1} + 3^B\left[ 1 + 1 \cdot 2 + 1 \cdot 2^2 ... + 1 \cdot 2^{C-1} \right]\right]$

Consider the expression:

$x_0 + x_1 \cdot 5 + x_2 \cdot 5^2 ... + x_{A-1} \cdot 5^{A-1} + 5^A\left[ y_0 + y_1 \cdot 3 + y_2 \cdot 3^2 ... + y_{B-1} \cdot 3^{B-1} + 3^B\left[ z_0 + z_1 \cdot 2 + z_2 \cdot 2^2 ... + z_{C-1} \cdot 2^{C-1} \right]\right]$

If we consider all combinations of $x_n, 0 \le n \le A-1$ to be any integer between $0$ and $4$ inclusive, $y_n, 0 \le n \le B-1$ to be any integer between $0$ and $2$ inclusive and $z_n, 0 \le n \le C-1$ to be any integer between $0$ and $1$ inclusive, we will generate all integers between $0$ and $N-1$ exactly once.

Now converting the above to a dice numbering we have

• $A$ 20-sided dice numbered $\left\{ 0, 5^n, 2 \cdot 5^n, 3 \cdot 5^n, 4 \cdot 5^n \right\}$ , each element repeated 4 times to fill up all 20 sides, for $0 \le n \le A-1$
• $B$ 6-sided dice numbered $\left\{ 0, 3^n 5^A, 2 \cdot 3^n 5^A \right\}$ , each element repeated 2 times to fill up all 6 sides, $0 \le n \le B-1$
• $C$ 4-sided dice numbered $\left\{ 0, 2^n 3^B 5^A \right\}$ , each element repeated 2 times to fill up all 4 sides, $0 \le n \le C-1$

However, the above dice numbering would pick an integer from $0$ to $N-1$ with equal probability, which still doesn't fit Vi's requirements. Luckily, there is a simple fix for that. We just have to add $1$ to all the sides of one of the die and we now generate an integer between $1$ and $N$ with equal probability. This leaves us with the construction shown above.

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We have shown 2 things:

1. If $N$ has prime factors that are not $2$ , $3$ or $5$ , then it is impossible
2. For every $N$ with prime factors $2, 3, 5$ , there is a dice numbering that satisfies Vi's requirements

With these 2 proven, we can safely conclude that it is possible if and only if $N$ only has prime factors $2$ , $3$ and $5$ .

From here, it is just a systematic count to get the answer $9825$ .

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I'm aware that this solution is very badly phrased. If you have any questions feel free to ask.