Discussion
Find the unit digit of \(4^{1} \times 9^{2} \times 4^{3} \times 9^{4} \times ... \times 4^{99} \times 9^{100}\)
Answer: D
The expression has 50 terms with base 4 and 50 terms with base 9. Let's group them: \((4^1 \times 4^3 \times ... \times 4^{99}) \times (9^2 \times 9^4 \times ... \times 9^{100})\). The unit digit of any odd power of 4 is 4. There are 50 such terms. The unit digit of any even power of 9 is 1. There are 50 such terms. So the unit digit is the unit digit of \((4 \times 4 \times ... (50 \text{ times})) \times (1 \times 1 \times ... (50 \text{ times}))\). This simplifies to the unit digit of \(4^{50} \times 1\). Since 50 is an even power, the unit digit of \(4^{50}\) is 6.
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