Question:medium

Four natural numbers selected at random are multiplied together, then the probability that the digit in the unit's place in the product be 1, 3, 7 or 9 is:

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This problem highlights a very neat shortcut condition: the set of digits $\{1, 3, 7, 9\}$ forms a closed group under multiplication (meaning multiplying any of these numbers together will always produce another number ending in 1, 3, 7, or 9). This guarantees that as long as individual numbers stay inside this group, the final product is guaranteed to stay inside it too!
Updated On: May 28, 2026
  • $\frac{16}{625}$
  • $\frac{18}{625}$
  • $\frac{4}{625}$
  • $\frac{5}{625}$
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
The unit's digit of a product depends only on the unit's digits of its factors. For a product to have a unit's digit of 1, 3, 7, or 9, the product must be odd and must not be a multiple of 5. This implies that none of the factors can be even and none of the factors can end in 5.
Step 2: Key Formula or Approach:
1. Identify the unit digits of individual factors that lead to the desired product units digit.
2. Probability \( P = (\text{Prob of one factor satisfies condition})^4 \).
Step 3: Detailed Explanation:
Possible unit digits for a natural number are \( \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\} \). Total = 10 digits.
Condition for the product to end in 1, 3, 7, or 9:
- The product is not divisible by 2 (all factors must be odd).
- The product is not divisible by 5 (none of the factors can end in 5).
So, the allowed unit digits for each factor are \( \{1, 3, 7, 9\} \).
Number of favorable unit digits for one number \( = 4 \).
Probability for one number to satisfy this condition \( = 4/10 = 2/5 \).
Since the four numbers are selected independently and multiplied:
\[ P(\text{Product ends in 1, 3, 7, 9}) = (2/5) \times (2/5) \times (2/5) \times (2/5) \]
\[ P = \frac{16}{625} \]
Step 4: Final Answer:
The probability is \( 16/625 \).
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