Discussion

Answer: D

We find the unit digit of each term separately.

Part 1: Unit digit of \(13^{25} \times 68^{57}\). This is the unit digit of \(3^{25} \times 8^{57}\). For \(3^{25}\), the cycle is (3,9,7,1) of length 4. \(25 \div 4\) gives rem 1, so unit digit is 3. For \(8^{57}\), the cycle is (8,4,2,6) of length 4. \(57 \div 4\) gives rem 1, so unit digit is 8. The unit digit of Part 1 is the unit digit of \(3 \times 8 = 24\), which is 4.

Part 2: Unit digit of \(24^{13} \times 57^{68}\). This is the unit digit of \(4^{13} \times 7^{68}\). For \(4^{13}\) (odd power), the unit digit is 4. For \(7^{68}\), the cycle is (7,9,3,1) of length 4. \(68\) is a multiple of 4, so the unit digit is the last in the cycle, which is 1. The unit digit of Part 2 is the unit digit of \(4 \times 1 = 4\).

The unit digit of the entire expression is the unit digit of the sum of the unit digits of the two parts: \(4 + 4 = 8\).