Question

\(in~~Z_{24},~~find~~x~~if~~x^n=0\)
please, help

Answer 1

@ganeshie8

Answer 2

wouldn't x=0 work ?

Answer 3

non trivial, please

Answer 4

I sense a disturbance in the force...

Answer 5

lol dumb q : what do we know about \(n\) ?

Answer 6

n in Z+

Answer 7

What is \(\Large \mathbb{Z}_{24}\) taken to be?
A ring?

Answer 8

yes, @Kurt

Answer 9

I have problem with the net... OMG.... I need know how to work out.... Please.. leave the guidance !!

Answer 10

...and, what? Are we looking for an element of \(\large \mathbb{Z}_{24}\) that is always zero when you raise it to any positive integer?

Answer 11

I think so

Answer 12

not any, some n

Answer 13

for some n? Okay, that simplifies things a bit.

Answer 14

Hmm.. @ganeshie8 ? :D

Answer 15

does this work ? \[x=2^a3^b\]
where \(a\ge 1\) and \(b\ge 1\)

Answer 16

yes it does, @ganeshie8

now... let's prove it! :D

Answer 17

yes, I think so, but what is the logic? I mean how to get that?

Answer 18

time for a little Kurt-magic? :)

haha, kidding :>

Answer 19

I have school now, but I do need know how to get it. Please. I appologize for not being here to get help.

Answer 20

btw, @ganeshie8
\[\Large x = 2^a \cdot 3^b \cdot \color{blue}k\]
seems to work, for all
\[\Large a,b,k \in \mathbb{Z}^+\]

Answer 21

that looks complete now :)

Answer 22

In English, L66, x must be divisible by both 2 and 3.

The plan has two phases, first, prove that if x is divisible by 2 and 3, then eventually, multiplying it by itself will yield a multiple of 24.

Second, show that IF x is indivisible by either 2 or 3, then it will never yield a multiple of 24 no matter how many times you multiply it by itself.

Answer 23

Okay, phase 1.

Suppose x is divisible by both 2 and 3.

That means that x is divisible by 6.

So... \(\large x = 6k\)

Then take n = 3.

\[\Large x^3 = 6^3k^3 = 216k^3 = 24\cdot 9\cdot k^3 \equiv 0 (\text{mod} \ 24)\]

Answer 24

Phase 2a:
Suppose x is odd. That is to say, it is indivisible by 2. It's simple enough to show that any integer power of an odd number will be odd. (Try to do that yourself: use induction!)

So, if x is odd, then x^n will also be odd, and therefore, we cannot have
\[\Large x^n = 24k = 2\cdot 12k \]

Answer 25

x^n will never be divisible by 24, no matter what positive integer value n takes.

Answer 26

Phase 2b requires you to show that if x is indivisible by 3, then it also cannot be raised to any power such that it will be divisible by 24. Use a similar approach to phase 2a.

Good luck!

Answer 27

And one non-essential hint that may well make your life easier:

All integers that are not divisible by 3 may be written as
\[\Large 3k \pm 1\]
Where k is an integer.

Best of luck :D

Answer 28

nice

Answer 29

???
if \(x^n\equiv 0\) (mod 24) , then \(24| x^n\)
24 =2*2*2*3
\(x^n =x*x*......*x~(ntimes)\)
hence 2*2*2*3|x*x*......*x
which leads to 2|x and 3|x or 6|x
the element x in \(Z_{24} \) satisfy x =6k are {0,6,12,18}
Does it work?