Four whole numbers taken at random are multiplied together. What is the chance that the last digit in the product is 1, 3, 7 or 9 ?

Options

A. `16/625`
B. `1/210`
C. `8/125`
D. `4/25`

Answer

To solve the problem of finding the probability that the last digit of the product of four randomly chosen whole numbers is 1, 3, 7, or 9, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Possible Last Digits**:
The last digit of a whole number can be any digit from 0 to 9. The digits we are interested in for the product to end in are 1, 3, 7, or 9.

2. **Count the Favorable Outcomes**:
The digits that can give us a last digit of 1, 3, 7, or 9 are:
- 1
- 3
- 7
- 9
Therefore, there are 4 favorable outcomes.

3. **Count the Total Outcomes**:
The total possible last digits (0 to 9) are 10.

4. **Calculate the Probability of Choosing One Digit**:
The probability of randomly selecting a digit that ends in 1, 3, 7, or 9 is:
\[ P(\text{last digit is 1, 3, 7, or 9}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{4}{10} = \frac{2}{5} \]

5. **Calculate the Probability for Four Digits**:
Since we are selecting four whole numbers, and we want all four to have a last digit of 1, 3, 7, or 9, we raise the probability to the power of 4:
\[ P(\text{all four digits end in 1, 3, 7, or 9}) = \left(\frac{2}{5}\right)^4 = \frac{16}{625} \]

6. **Final Result**:
Therefore, the probability that the last digit of the product of four randomly chosen whole numbers is 1, 3, 7, or 9 is:
\[ \frac{16}{625} \]

### Conclusion:
The final answer is \(\frac{16}{625}\).