I have a \(2^1\)-sided dice with sides labelled \( 1,2 \), a \(2^2\)-sided dice with sides labelled \( 1,2,3, 4 \), and so on, up to a \(2^n\)-sided dice with sides labelled \( 1,2,3,...,2^n \). I simultaneously roll all \(n\) of them. Let \( d_{i} \) denote the number that comes up on the dice with \( i \) sides.

How many ways can I roll the dice such that

$3 \mid \displaystyle\sum_{i=1}^{n} d_{i}$

- What is the answer when $n=1$?
- What is the answer when $n = 2$?
- Tabulate the initial values for small $n$.
- What is your guess for the answer for general $n$?
- How would you prove it?
- Find another way of proving this.
- If possible, generalize to odd $m$:

$m \mid \displaystyle\sum_{i=1}^{n} d_{i}$

This note is inspired by Josh Rowley's Dice Sums 3.

A simple solution existed because there was a dice with $s_i3$ sides, and $m=3$ divides that value. In this case, we do not have $m \mid s_i =2^i$.

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## Comments

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TopNewest@Josh Rowley Any other interesting variants on dice sums?

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I was thinking something along the lines of: $n$ dice, where dice $i$ has $i$ sides labelled $1,2,3,...,i$ for all $1 \le i \le n$ (ie. a standard set up). Then, given that $3 \mid \displaystyle\sum_{i=1}^{n}$ What is the expected number of dice which show the number 1? Unfortunately I'm having problems answering it myself...

ps in the problem you've posted I think it should state $3 \mid \displaystyle\sum_{i=1}^{n} d_{2^i}$ instead of $3 \mid \displaystyle\sum_{i=1}^{n} d_{i}$

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Re notation, my $i$th dice has $2^i$ sides, so the notation is fine. Though, it might be good to keep notation consistent, so that it's easier to talk about more general cases.

I have a conjecture for the case of rolling $i$ dice labelled with $1, 2, 3, \ldots, s_i$, and the number of ways to get a sum equal to $k \pmod{n}$.

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CAN i=1 you have written i=or >1,,,

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How can you have a 2-sided die?

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A 2-sided die is essentially a coin :)

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The case for and odd $m$ i think is the same as that of 3. The roll all the dice except m.And whatever the outcome may be there is only one possible choice of the $m$th die so that the whole sum is divisible by $m$.So the total number of possibilities = $(2^{n})!/m$

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Note the change in my notation. Each of the dice has sides of $2^i$, and so in particular $m \not \mid 2^i$.

Also, there is the question of what to do when $m > n$ in Josh's original problem. E.g. think of having 40 dice, and you want the sum to be divisible by 41 (or 43). In this case, $2^{40}! / 41$ is not an integer.

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Oh...I missed the point that powers of 2 were being used ...and yes in the original problem also it will be a difficult case when $m>n$.........

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I think we have to find a bijection for the case where $m>n$

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