Let m=77^{n}. The index n is given a positive integral value at random. What is the probability that the value of m will have 1 in the units place?
- A. \frac{1}{2}
- B. \frac{1}{3}
- C. \frac{1}{4} ✓
- D. \frac{1}{n}
Correct Answer: C. \frac{1}{4}
Explanation
The unit digit of 77^n is determined by 7^n. The unit digits of powers of 7 follow the cycle: 7^1 \rightarrow 7, 7^2 \rightarrow 9, 7^3 \rightarrow 3, 7^4 \rightarrow 1. This is a cycle of length 4. The unit digit is 1 only when n is a multiple of 4. Assuming n takes positive integral values uniformly, the probability is \frac{1}{4}.
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