Let `x = 33^n` . The index n is given a positive integral value at random. The probability that the value of x will have 3 in the units place is

Options

A. `1/3`
B. `1/4`
C. `1/5`
D. `1/2`

Answer

To solve the problem of finding the probability that the value of \( x = 33^n \) has 3 in the units place, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression \( x = 33^n \). We need to determine the units digit of \( x \) for different values of \( n \).

2. **Identifying the Units Digit**:
The units digit of \( 33^n \) can be determined by looking at the units digit of the base, which is 3. Therefore, we need to find the units digit of \( 3^n \).

3. **Finding the Pattern in Units Digits of Powers of 3**:
Let's calculate the units digits of the first few powers of 3:
- \( 3^1 = 3 \) (units digit is 3)
- \( 3^2 = 9 \) (units digit is 9)
- \( 3^3 = 27 \) (units digit is 7)
- \( 3^4 = 81 \) (units digit is 1)
- \( 3^5 = 243 \) (units digit is 3)
- \( 3^6 = 729 \) (units digit is 9)
- \( 3^7 = 2187 \) (units digit is 7)
- \( 3^8 = 6561 \) (units digit is 1)

From this, we can see that the units digits repeat every 4 terms: \( 3, 9, 7, 1 \).

4. **Determining When the Units Digit is 3**:
The units digit of \( 3^n \) is 3 when \( n \equiv 1 \mod 4 \). This means that \( n \) can take the values \( 1, 5, 9, 13, \ldots \) (i.e., \( n = 4k + 1 \) for \( k \in \mathbb{Z}_{\geq 0} \)).

5. **Calculating the Probability**:
Since the units digits repeat every 4 values of \( n \), we can conclude that:
- Out of every 4 consecutive integers, there is exactly 1 integer (specifically those of the form \( 4k + 1 \)) that results in a units digit of 3.

Therefore, the probability \( P \) that \( x \) has a units digit of 3 is given by:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1}{4} \]

### Final Answer:
The probability that the value of \( x = 33^n \) will have 3 in the units place is \( \frac{1}{4} \).