Math Series #5

Let the given number be $N$ . Then

$\begin{aligned} N & \equiv 18^{205} - 42^{94} \pmod {10} \\ & \equiv 8^{205} - 2^{94} \pmod {10} \\ & \equiv (10-2)^{205} - 2^{94} \pmod {10} \\ & \equiv -2^{205} - 2^{94} \pmod {10} \\ & \equiv -2 \times 2^{4 \times 51} - 2^2 \times 2^{4 \times 23} \pmod {10} \\ & \equiv -2 \times 16^{51} - 2^2 \times 16^{23} \pmod {10} \\ & \equiv -2 \times \blue{6^{51}} - 2^2 \times \blue{6^{23}} \pmod {10} & \small \blue{\text{Note that any positive power of 6 always ends with 6.}} \\ & \equiv - 2 \times 6 - 4 \times 6 \pmod {10} \\ & \equiv - 12 - 24 \equiv - 6 \equiv \boxed 4 \pmod {10} \end{aligned}$

We just have to consider the powers of last digits of each term. If we work out the last digits of the power of last digit of the base of a term, we will find recurrence.
[The pattern for last digit depends on the power and last digit of base because the last digit comes from the multiplication of the base and the previous power. And among the digits of the base only the last digit is responsible for the last digit of product.]

If we work out for powers of $8$ we see $8, 4, 2, 6, 8, 4, 2,....$ and so on as the last digits for power $1, 2, 3,....etc$ . It repeats after $4$ terms. Now we can write $205=4*51+1$ . So last digit of $18^{205}$ is equal to the last digit of $8^1$ which is $8$ .

Similarly for the $2nd$ term, last digit of powers of $2$ also repeats after $4$ terms. Now, $394=4*98+2$ . So last digit of $42^{394}$ is equal to the last digit of $2^2$ which is $4$ .

So, the last digit is $8-4=4$ .