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Number Theory

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Easy

Fermat Number Last Digit

NUMTH-UZBF1B

A Fermat's Number is a positive integer of the form $2^{2^n}+1$ for some nonnegative integer $n$.

The first few numbers are $3$, $5$, $17$, $257$, $65537$, $4294967297$, $\cdots$

We see that except for the first two, all Fermat's Numbers have last digit $7$. The following is a proof of this fact. Fill in the blanks in the proof.

Proof

We consider $2^k$ modulo $10$. We have the first four

$$ 2^1 \equiv 2 \ \mathrm{mod} \ 10, \ \ 2^2 \equiv 4 \ \mathrm{mod} \ 10, \ \ 2^3 \equiv 8 \ \mathrm{mod} \ 10, \ \ 2^4 \equiv 6 \ \mathrm{mod} \ 10 $$

Since $2^5 \equiv 2 \ \mathrm{mod} \ 10$, we see that a pattern $2$, $4$, $8$, $\underline{...}$ repeats in $2^k \ \mathrm{mod} \ 10$.

We have $2^n \equiv 0 \ \mathrm{mod} \ 4$ if $n\underline{...}$. Thus, we obtain that

$$ 2^{2^n} \equiv \underline{...} \ \mathrm{mod} \underline{...} $$

if $n\underline{...}$

Therefore, we have

$$ 2^{2^n}+1 \equiv \underline{...} \ \mathrm{mod} \underline{...} $$

for $n\underline{...}$

A

$6$, $\geq 2$, $6$, $4$, $\geq 2$, $6$, $4$, $\geq 2$.

B

$6$, $\geq 1$, $6$, $10$, $\geq 1$, $7$, $10$, $\geq 1$.

C

$6$, $\geq 2$, $6$, $10$, $\geq 2$, $7$, $10$, $\geq 2$.

D

$6$, $\geq 2$, $6$, $10$, $\geq 2$, $7$, $10$, $\geq 1$.

E

$4$, $\geq 2$, $4$, $10$, $\geq 2$, $5$, $10$, $\geq 2$.