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

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

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 $m \mid s_i =2^i$ .

Image credit: Wikipedia Matěj Baťha

#Combinatorics #MathematicalInduction #RootsOfUnity #PatternRecognition #GeneratingFunctions

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

@Josh Rowley Any other interesting variants on dice sums?

Calvin Lin Staff - 7 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 $3 \mid \displaystyle\sum_{i=1}^{n} d_{i}$

Josh Rowley - 7 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 $k \pmod{n}$ .

Calvin Lin Staff - 7 years, 1 month ago

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

Rohit Singh - 7 years, 1 month ago

How can you have a 2-sided die?

Frodo Baggins - 7 years, 1 month ago

A 2-sided die is essentially a coin :)

Eddie The Head - 7 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 = $(2^{n})!/m$

Eddie The Head - 7 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.

Calvin Lin Staff - 7 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 $m>n$ .........

Eddie The Head - 7 years, 1 month ago

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

Eddie The Head - 7 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$