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 \( m \mid s_i =2^i\).

## Comments

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TopNewest@Josh Rowley Any other interesting variants on dice sums? – Calvin Lin Staff · 2 years, 11 months ago

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ps in the problem you've posted I think it should state \( 3 \mid \displaystyle\sum_{i=1}^{n} d_{2^i} \) instead of \( 3 \mid \displaystyle\sum_{i=1}^{n} d_{i} \) – Josh Rowley · 2 years, 11 months ago

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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 \( k \pmod{n} \). – Calvin Lin Staff · 2 years, 11 months ago

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– Rohit Singh · 2 years, 11 months ago

CAN i=1 you have written i=or >1,,,Log in to reply

How can you have a 2-sided die? – Frodo Baggins · 2 years, 11 months ago

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– Eddie The Head · 2 years, 11 months ago

A 2-sided die is essentially a coin :)Log in to reply

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 = \((2^{n})!/m\) – Eddie The Head · 2 years, 11 months ago

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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. – Calvin Lin Staff · 2 years, 11 months ago

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– Eddie The Head · 2 years, 11 months ago

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

– Eddie The Head · 2 years, 11 months ago

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 \(m>n\).........Log in to reply