NT problem

I'm assuming that the expession ((2^1993 + 3^1993) is a multiple of 5.

(2^1993 + 3^1993)/5  mod 10^10==1964726983 - these are the last 10 digits of the quotient.

If you can determine the last two digits of $2^{1993} + 3^{1993}$, it suffices to determine the last digit of $\dfrac{2^{1993} + 3^{1993}}5$.

$2^{1993} + 3^{1993}$ is divisible by 5 because $2^{1993} + 3^{1993} \equiv 2^{1993} + (-2)^{1993} \equiv 2^{1993} - 2^{1993} \equiv 0 \pmod 5$.

Note that $2^{1993} \equiv 2^{1993 \operatorname{mod}40}\equiv 2^{33} \pmod{100}$  by Euler's theorem.

Also $2^{33} \equiv 0 \pmod 4$ and $2^{33} \equiv 2^{33\operatorname{mod}20} \equiv 2^{13} \equiv 8192 \equiv 17\pmod {25}$ again by Euler's theorem. 

By Chinese remainder theorem, $2^{1993} \equiv 2^{33} \equiv 92 \pmod {100}$.

Similarly with $3^{1993}$, we have $3^{1993} \equiv 23\pmod{100}$

Therefore, $2^{1993} + 3^{1993} \equiv 92 + 23 \equiv 115 \equiv 15 \pmod{100}$.

Then $2^{1993} + 3^{1993} = 100k + 15$ for some integer k.

Then $\dfrac{2^{1993} + 3^{1993}}5 = 20k + 3$ for some integer k. Therefore, the last digit is 3.