<s>[INST] 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?
 [/INST] 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.
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