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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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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I think we have to find a bijection for the case where \(m>n\)

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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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