If n is a positive integer then the probability that `3^(n)` has 3 at unit place is

Options

A. `1/4`
B. `1/5`
C. `3/10`
D. `1/8`

Answer

To find the probability that \(3^n\) has 3 at the unit place when \(n\) is a positive integer, we can follow these steps:

### Step 1: Identify the pattern of unit digits for powers of 3
Let's calculate the unit digits of the first few powers of 3:
- \(3^1 = 3\) (unit digit is 3)
- \(3^2 = 9\) (unit digit is 9)
- \(3^3 = 27\) (unit digit is 7)
- \(3^4 = 81\) (unit digit is 1)
- \(3^5 = 243\) (unit digit is 3)
- \(3^6 = 729\) (unit digit is 9)
- \(3^7 = 2187\) (unit digit is 7)
- \(3^8 = 6561\) (unit digit is 1)

### Step 2: Observe the repeating pattern
From the calculations above, we can see that the unit digits repeat every 4 powers:
- The sequence of unit digits is: 3, 9, 7, 1.

### Step 3: Determine the probability
Since the unit digits repeat every 4 powers, we can conclude:
- The unit digit is 3 for \(n \equiv 1 \mod 4\) (i.e., when \(n\) is of the form \(4k + 1\) for \(k \geq 0\)).
- The total number of distinct outcomes for the unit digits is 4 (3, 9, 7, 1).

Thus, the probability that \(3^n\) has 3 at the unit place is:
\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1}{4} \]

### Final Answer:
The probability that \(3^n\) has 3 at the unit place is \(\frac{1}{4}\).

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