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Dice Sums 3+

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

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

1. What is the answer when $$n=1$$?
2. What is the answer when $$n = 2$$?
3. Tabulate the initial values for small $$n$$.
4. What is your guess for the answer for general $$n$$?
5. How would you prove it?
6. Find another way of proving this.
7. 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 $$m \mid s_i =2^i$$.

Image credit: Wikipedia Matěj Baťha

Note by Calvin Lin
3 years, 1 month ago

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@Josh Rowley Any other interesting variants on dice sums? Staff · 3 years, 1 month ago

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 $$3 \mid \displaystyle\sum_{i=1}^{n} d_{i}$$ · 3 years, 1 month ago

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 $$k \pmod{n}$$. Staff · 3 years, 1 month ago

CAN i=1 you have written i=or >1,,, · 3 years, 1 month ago

How can you have a 2-sided die? · 3 years, 1 month ago

A 2-sided die is essentially a coin :) · 3 years, 1 month ago

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 = $$(2^{n})!/m$$ · 3 years, 1 month ago

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. Staff · 3 years, 1 month ago

I think we have to find a bijection for the case where $$m>n$$ · 3 years, 1 month 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 $$m>n$$......... · 3 years, 1 month ago

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