How to denote all unrepeated combinations of numbers in a set?

Question

In my homework assignment, there's a task:

2. Find set of all numbers which can divide 24.

Translation mine, I hope it's correct.

Through integer factorization I know that 24 is 2*2*2*3. Now I could just manually do all combinations of these four numbers and write them in a set:

$$\{2, 2*2, 2*2*2, 2*3, 2*3*2, \dots\}$$

That's a lot of work (or would be for bigger numbers), but maybe not the kind of work the homework expects me to do. Is there a way to denote a set of combinations of numbers? Basically:

Let A be a set of numbers such that it contains all combinations of 2, 2, 2 and 3 without repetition.

How to write that as math?

Answer 1

If you mean without repetition, you would just say all subsets of $\{2,3\}$. Maybe you want to exclude the empty set. That will not get you the list you are showing, because you show many cases with more than one $2$. You could note that the factors of $24$ have zero to three factors of $2$ and zero or one factor of $3$. Do you include $1$, which has no factors of either, and $24$, which is not a proper factor? If so, you set can be written as $$\{2^a3^b\mid 0 \le a \le 3, 0 \le b \le 1\}$$

Answer 2

I don't know what level of class your homework is for but there are two possible answers.

1) is simply to list all the values: {1,2,3,4,6,8,12,24}. I don't think that can be considered too much work.

2) You list them as {$2,2*2,2*2*2,2*3,2*2*3,2*2*3$} [sic] which you note is tedious (although you actually had only one more to list). When you say "Let A be a set of numbers such that it contains all combinations of 2, 2, 2 and 3 without repetition" you actually "Let A be a set of numbers whose prime factors are 2 and 3".

Use exponents.

{$2,2^2,2^3,2*3,2^2*3,2^3*3$} [sic]. To write this in math: $A =\{2^a*3^b|0\le a \le 3; 0 \le b \le 1\}$. Notice that this includes both $1$ and $24$ as numbers that divide $24$ which you omitted.

Now to list these without repetition and worries of "where was I; what's next", simply list the powers of twos, from 1 to 8 first, then 3 times the powers of twos in order.

$A = \{1,2,4,8,$ pause $3,6,12,24\}$

So if you wanted to find all the factors of $1800=8*9*25=2^33^25^2$ for example you would list:

$A=\{1,2,4,8$ --> in other words that is $2^0$ to $2^3$

$,3,6,12,24$ pause $,9,18,36,72$ ---> in other words, that is $3*2^0$ to $3*2^3$ and then $3^2*2^0$ to $3^2*2^3$

$5,10,20,40,15,30,60,120,45,90,180,360$--> in other words, that is $5$ times everything done so so

$25,50,100,200,75,150,300,600,225,450,900,1800\}$ ---> in other words that is $5^2$ times everything done so far.

$= \{2^a*3^b*5^c|0 \le a \le 3; 0 \le b \le 2; 0 \le c \le 2\}$