How big!

Method 1:

$2074^{35849} \equiv (2070+4)^{35849} \equiv 4^{35849} \text{ (mod 10)}$ .

Now, we note that $4^n \equiv \begin{cases} 4 \text{ (mod 10)} & \text{if }n \text{ is odd.} \\ 6 \text{ (mod 10)} & \text{if }n \text{ is even.} \end{cases}$ .

Therefore, $2074^{35849} \equiv \boxed{4} \text{ (mod 10)}$ .

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Method 2:

We need to find $2074^{35849} \text{ mod }10$ . Since 2074 and 10 are not coprime integers, we have to consider $2074^{35849} \text{ mod }2$ and $2074^{35849} \text{ mod }5$ separately.

$\begin{aligned} 2074^{35849} & \equiv 0 \text{ (mod 2)} \\ 2074^{35849} & \equiv 2074^{\color{#3D99F6}35849 \text{ mod }\phi(5)} \text{ (mod 5)} & \small \color{#3D99F6} \text{Since } \gcd (2074, 5) = 1 \text{, Euler's theorem applies.} \\ & \equiv 2074^{\color{#3D99F6}35849 \text{ mod }4} \text{ (mod 5)} & \small \color{#3D99F6} \text{Euler totient function }\phi (5) = 4 \\ & \equiv 2074^1 \equiv 4 \text{ (mod 5)} \\ \implies 5 {\color{#3D99F6}n} +4 & \equiv 0 \text{ (mod 2)} & \small \color{#3D99F6} \text{where }n \text{ is an integer.} \\ \implies n & \equiv 0 \\ \implies 2074^{35849} & \equiv 5(0) + 4 \equiv \boxed{4} \text{ (mod 10)} \end{aligned}$

We know that,

$4^1 = 4$

$4^2 = 16$

$4^3 = 64$

$4^4 = 256$

Observed that the last digit repeats in every cycle of $2$ , that means that we must divide the exponent by $2$ . If there is no remainder, the last digit is $6$ . If the remainder is $1$ the last digit is $4$ .

Dividing $35849$ by $2$ gives a remainder of $1$ , so the last digit is $\boxed{4}$ .

2074 = 4 mod 10

Now $2074 ^2$ = 6 mod 10

Again $2074 ^3$ = 4 mod 10

Hence for odd n last digit is $\boxed{4}$