There exist a set $X$ such that the number of function $y: x\to \{1,2,3\}$ is $1000$. There exist a set $X$ such that the number of function $y: x\to \{1,2,3\}$ is $1000.$
My attempt:
False, Let set $z = \{1,2,3\}$ then $|z|^{|x|}$ is set of function $y:x\to z.$
$|x| = n$ and $|z| = \{1,2,3\} = 3$. Thus $|z|^{|x|} = 3^n$.
$3^1 \neq 1000,~ 3^2 \neq 1000,~ 3^3 \neq 1000$. Thus $3^n \neq 1000$.
Can anyone please give feedback on my answer and tell me whether the solution is correct or no?

There is an old joke about the ways that people in different
professional fields "prove" that all odd numbers greater than $2$ are prime.
One way is, "$3$ is a prime, $5$ is a prime, $7$ is a prime, therefore
all odd numbers greater than $2$ are prime."
You do not want people thinking your reasoning was like that.
There are a few ways to show that there is no $n$ such that $3^n$ is $1000.$
Prime factorization theorem (already used by Hagen von Eitzen).
Every positive integer has a unique factorization as powers of
positive prime numbers. For any $n$, $3^n$ has a prime factor of $3$,
therefore $3 \mid 3^n$. But $1000 = 2^3\cdot5^3$ has no factor of $3$,
so it cannot equal $3^n$ for any $n$.
Arithmetic modulo $m$, or in this case, "what is the last digit."
Any positive integer is congruent (modulo $10$) to its last digit.
In particular,
$$3^1 \equiv 3 \pmod{10}$$
$$3^2 \equiv 9 \pmod{10}$$
$$3^3 = 27 \equiv 7 \pmod{10}$$
$$3^4 = 81 \equiv 1 \pmod{10}$$
After that, successive multiplication by $3$ repeats the pattern $3,9,7,1$
indefinitely. You never see $0$ as a last digit, so $1000$ cannot be $3^n$.
Exhaustive search. You were part way there, but did not go far enough.
$$3^1 = 3 \neq 1000$$
$$3^2 = 9 \neq 1000$$
$$3^3 = 27 \neq 1000$$
$$3^4 = 81 \neq 1000$$
$$3^5 = 243 \neq 1000$$
$$3^6 = 729 \neq 1000$$
$$3^7 = 2187 \neq 1000$$
For $n>7,$ $3^n > 3^7 > 1000$, so $3^n \neq 1000.$
For this particular kind of problem, prime factorization looks like the
clear winner to me.
