Natural numbers of the form $F_n=2^{2^n} + 1 $ are called Fermat numbers. In 1640, Fermat conjectured that all numbers $F_n$, where $n\neq 0$, are prime. (The conjecture was later shown to be false.) What is the units digit of $F_{1000}$?
Let's first try to find the cycle of units digits of $2^n$, starting with $n=1$: $2, 4, 8, 6, 2, 4, 8, 6,\ldots$ . The cycle of units digits of $2^n$ is 4 digits long: 2, 4, 8, 6. To find the units digit of $2^n$, for any positive integer $n$, we simply need to find the remainder, $R$, when $n$ is divided by 4 ($R=1$ corresponds to the units digit 2, $R=2$ corresponds to the units digit 4, etc.) Since $2^{1000}\div4=2^{998}$ without remainder, the units digit of $2^{2^{1000}}$ is 6. Therefore, the units digit of $F_n=2^{2^{1000}}+1$ is $6+1=\boxed{7}$.
7
Level 5
Number Theory
Problem: Natural numbers of the form $F_n=2^{2^n} + 1 $ are called Fermat numbers. In 1640, Fermat conjectured that all numbers $F_n$, where $n\neq 0$, are prime. (The conjecture was later shown to be false.) What is the units digit of $F_{1000}$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the units digit of $F_{1000} = 2^{2^{1000}} + 1$, we first need to determine the units digit of $2^{2^{1000}}$. The units digits of powers of 2 follow a repeating cycle: 2, 4, 8, 6. This cycle has a length of 4. 

Next, we need to find $2^{1000} \mod 4$ to determine which number in the cycle corresponds to $2^{2^{1000}}$. Since $2^{1000}$ is a power of 2 greater than or equal to $2^2$, we have $2^{1000} \equiv 0 \mod 4$. 

Now, we can use this result to find the position in the cycle. The cycle of units digits for powers of 2 is indexed as follows:
- $2^1 \equiv 2$ (units digit 2)
- $2^2 \equiv 4$ (units digit 4)
- $2^3 \equiv 8$ (units digit 8)
- $2^4 \equiv 16$ (units digit 6)

Since $2^{1000} \equiv 0 \mod 4$, we look at the 4th position in the cycle, which corresponds to the units digit of $2^4$, giving us a units digit of 6 for $2^{2^{1000}}$.

Finally, we add 1 to this result to find the units digit of $F_{1000}$: 
6 (units digit of $2^{2^{1000}}$) + 1 = 7.

Thus, the final answer is 7.<|END_OF_LATENT|> The final answer is $7$.