Question

Amy is playing a board game and rolls two number cubes. Let A = {the sum of the number cubes is odd}, and let B = {the sum of the number cubes is divisible by 3}. List the outcomes in A ∪ B.

Answer 1

it is a lot

Answer 2

saying number cubes instead of dice? lawl i smell trick question....
nonetheless... they ARE dice, right? ^.^

Answer 3

old heads call them dice
also normal people
scared math teachers who are afraid of being called satanic gamblers call them "number cubes" which frankly sounds a lot more cult like than "dice" to me

Answer 4

all i need to know is that they're both marked 1 through 6, and we can get this thing rolling... quite literally, actually :D

Answer 5

there are a lot that have a sum of an odd number
half of the 36 possible combinations are odd
you have to list all 18

Answer 6

would it be {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

Answer 7

then also list the even ones that are divisible by 3
unfortunately no, it is not that brief

Answer 8

no, silly, your list should contain only outcomes divisible by three, or outcomes that are even...

Answer 9

Oh okay

Answer 10

and second, yeah, it's probably not that simple... you might have to record the individual outcomes of each <ehem> number cube.

Answer 11

3, 6, 9, 12 all divisble by 3

Answer 12

also outcomes look like ordered pairs
one for each die
like for example \((1,2)\) and \((2,1)\) the two ways to get a total of 3

Answer 13

let me see if i can find a dice table to look at

Answer 14

\[ \left[\begin{matrix}\left(\begin{matrix}\color{red}1&\color{blue}1\\\color{red}1&\color{blue}2\\\color{red}1&\color{blue}3\\\color{red}1&\color{blue}4\\\color{red}1&\color{blue}5\\\color{red}1&\color{blue}6\end{matrix}\right)&\left(\begin{matrix}\color{red}2&\color{blue}1\\\color{red}2&\color{blue}2\\\color{red}2&\color{blue}3\\\color{red}2&\color{blue}4\\\color{red}2&\color{blue}5\\\color{red}2&\color{blue}6\end{matrix}\right)&\left(\begin{matrix}\color{red}3&\color{blue}1\\\color{red}3&\color{blue}2\\\color{red}3&\color{blue}3\\\color{red}3&\color{blue}4\\\color{red}3&\color{blue}5\\\color{red}3&\color{blue}6\end{matrix}\right)&\left(\begin{matrix}\color{red}4&\color{blue}1\\\color{red}4&\color{blue}2\\\color{red}4&\color{blue}3\\\color{red}4&\color{blue}4\\\color{red}4&\color{blue}5\\\color{red}4&\color{blue}6\end{matrix}\right)&\left(\begin{matrix}\color{red}5&\color{blue}1\\\color{red}5&\color{blue}2\\\color{red}5&\color{blue}3\\\color{red}5&\color{blue}4\\\color{red}5&\color{blue}5\\\color{red}5&\color{blue}6\end{matrix}\right)&\left(\begin{matrix}\color{red}6&\color{blue}1\\\color{red}6&\color{blue}2\\\color{red}6&\color{blue}3\\\color{red}6&\color{blue}4\\\color{red}6&\color{blue}5\\\color{red}6&\color{blue}6\end{matrix}\right)\end{matrix}\right]\]

Answer 15

now all you have to do is peck through these thirty-six possible outcomes and pick out the ones that give you either an odd sum or a divisible-by-three sum.

good luck XD