+9673 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.
