# 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
6 years, 7 months ago

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@Josh Rowley Any other interesting variants on dice sums?

Staff - 6 years, 7 months 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 $3 \mid \displaystyle\sum_{i=1}^{n} d_{i}$

- 6 years, 7 months 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 - 6 years, 7 months ago

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

- 6 years, 7 months ago

How can you have a 2-sided die?

- 6 years, 7 months ago

A 2-sided die is essentially a coin :)

- 6 years, 7 months 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 = $(2^{n})!/m$

- 6 years, 7 months 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 - 6 years, 7 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 $m>n$.........

- 6 years, 7 months ago

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

- 6 years, 7 months ago