Number theory

Easy Solution:

This solution is for a more easier approach other than using modulo.

First, instead of actually solving $3^{1999}$ and $2^{1999}$ , we can find the last digit of both of these two large numbers by finding a pattern.

First, lets find the last digit when the exponent is 0, then 1, then 2 and so on:

$3^{0}$ = 1; $2^{0} = 1$

$3^{1}$ = 3; $2^{1} = 2$

$3^{2}$ = 9; $2^{2} = 4$

$3^{3}$ = 7; $2^{3} = 8$

$3^{4}$ = 1; $2^{4} = 6$

$3^{5}$ = 3; $2^{5} = 2$

At this point, we could wonder if this reoccurring pattern will happen every 4th power. We can check it by finding out the next time the last digit will be equivalent to the first number.

5 + 4 = 9

$3^{9}$ = 3; $2^{9}$ = 2

Now, we can assume that the pattern happens every 4th power. Utilizing the pattern, the problem would be solve very easily.

$3^{1999}$ = $3 \times 3^{2}$ ; $2^{1999}$ = $2 \times 2^{2}$

$3^{1999}$ = $\boxed{7}$ ; $2^{1999}$ = $\boxed{8}$

7 + 8 = 15; last digit = $\boxed{5}$

Let us consider $3^{1999} \text{ mod 10}$ and $2^{1999} \text{ mod 10}$ separately.

$\begin{aligned} 3^{1999} & \equiv 3^{\color{#3D99F6}{1999 \text{ mod }\phi (10)}} \text{ (mod 10)} & \small \color{#3D99F6}{\text{Since }\gcd(3,10) = 1 \text{, Euler's theorem applies.}} \\ & \equiv 3^{\color{#3D99F6}{1999 \text{ mod }4}} \text{ (mod 10)} \\ & \equiv 3^{\color{#3D99F6}{3}} \text{ (mod 10)} \\ & \equiv 7 \text{ (mod 10)} \end{aligned}$

Now consider $2^{1999} \text{ mod 2}$ and $2^{1999} \text{ mod 5}$ separately.

$\begin{aligned} 2^{1999} & \equiv 0 \text{ (mod 2)} \\ 2^{1999} & \equiv 2^{\color{#3D99F6}{1999 \text{ mod }\phi (5)}} \text{ (mod 5)} & \small \color{#3D99F6}{\text{Since }\gcd(2,5) = 1 \text{, Euler's theorem applies.}} \\ & \equiv 2^{\color{#3D99F6}{1999 \text{ mod }4}} \text{ (mod 5)} \\ & \equiv 2^{\color{#3D99F6}{3}} \text{ (mod 5)} \\ & \equiv 3 \text{ (mod 5)} \\ \implies 2^{1999} & \equiv 8 \text{ (mod 10)} \end{aligned}$

Therefore, we have:

$\begin{aligned} 3^{1999} + 2^{1999} & \equiv 7 + 8 \text{ (mod 10)} \\ & \equiv \boxed{5} \text{ (mod 10)} \end{aligned}$