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)/(2)`
B. `(1)/(3)`
C. `(1)/(4)`
D. none of these

Explanation

To solve the problem, we need to determine the probability that the value of \( x = 33^n \) has 3 in the units place when \( n \) is a positive integer.

### Step-by-Step Solution:

1. **Identify the Units Digit of Powers of 33**:
We first need to find the units digit of \( 33^n \) for different values of \( n \). The units digit of a number is determined by the units digit of its base raised to the power.

- For \( n = 1 \):
\[ 33^1 = 33 \quad \text{(units digit is 3)} \]
- For \( n = 2 \):
\[ 33^2 = 1089 \quad \text{(units digit is 9)} \]
- For \( n = 3 \):
\[ 33^3 = 35937 \quad \text{(units digit is 7)} \]
- For \( n = 4 \):
\[ 33^4 = 1185921 \quad \text{(units digit is 1)} \]

2. **Observe the Pattern**:
We can observe that the units digits repeat every 4 powers:
- \( n = 1 \): 3
- \( n = 2 \): 9
- \( n = 3 \): 7
- \( n = 4 \): 1

Thus, the sequence of units digits for \( 33^n \) is: 3, 9, 7, 1.

3. **Determine the Cycle**:
The pattern of units digits (3, 9, 7, 1) repeats every 4 values of \( n \). Therefore, we can conclude that:
- For \( n \equiv 1 \mod 4 \), the units digit is 3.
- For \( n \equiv 2 \mod 4 \), the units digit is 9.
- For \( n \equiv 3 \mod 4 \), the units digit is 7.
- For \( n \equiv 0 \mod 4 \), the units digit is 1.

4. **Calculate the Probability**:
Out of the 4 possible outcomes (3, 9, 7, 1), only one outcome (3) corresponds to having 3 in the units place. Therefore, the probability \( P \) that the units digit of \( 33^n \) is 3 is given by:
\[ P(\text{units digit is 3}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1}{4} \]

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

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