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The last digit of $2^{80}$ is..

1. 2
2. 4
3. 6
4. 8

 $2^1$ Last digit is $2$ $1 \mod 4 = 1$ $2^2$ Last digit is $4$ $2 \mod 4 = 2$ $2^3$ Last digit is $8$ $3 \mod 4 = 3$ $2^4$ Last digit is $6$ $0 \mod 4 = 0$ $2^5$ Last digit is $2$ $1 \mod 4 = 1$ $\dots$ $\dots$

so we see that there is some repetition after a occurrence of 4.
so $2^{80}$ where $80 \ mod \ 4 = 0$, Hence last digit is 6

with as number only 4 possible unit digit is possible

21 = 2     4n+1       remainder 1    unit digit 2
22 = 4     4n+2       remainder 2    unit digit 4
23 = 8     4n+3       remainder 3    unit digit 8
24 = 16   4n           remainder 0     unit digit 6
after this unit digit will repeat so check the remainder of power%4

here 280 therefore 80%4 = 0 so unit digit will be 6 Option C

To find the last digit of $2^{80}$, we need to find $2^{80} \text{ mod }10$.

Note that $\text{gcd}(2^{80}, 10)=2$, and $m \cdot a \text{ mod } m \cdot n = m (a \text{ mod } n)$. Then $2^{80} \text{ mod }10 = 2 \cdot (2^{79} \text{ mod }5)$.

Also, since $\text{gcd}(2,5)=1$ and $\phi(5)=4$, where $\phi$ is the Euler's Totient function, we have $2^4 \text{ mod } 5 =1$, by Euler's Totient Theorem.

Since $2^4 \text{ mod } 5 =1$, we have $2^{79} \text{ mod }5 = ((2^4)^{19} \cdot 2^3 ) \text{ mod }5 = 1\cdot3=3$.

Thus, $2^{80} \text{ mod }10 = 2 \cdot (2^{79} \text{ mod }5) = 2 \cdot 3 =6$, which is the last digit. Therefore, the answer is Option C.

Similarly, to calculate last two digits we need $2^{80} \text{ mod }100$, for last three digits we need $2^{80} \text{ mod }1000$, and so on.