Last digit behavior under multiplication: Four whole numbers are chosen independently at random. What is the probability that the last digit of their product is in {1, 3, 7, 9}?

Difficulty: Medium

Correct Answer: 16/625

Explanation:


Introduction / Context:
The last digit of a product depends only on the last digits of the factors modulo 10. A product will end in 1, 3, 7, or 9 precisely when it has no factor 2 or 5 in its decimal factorization, i.e., every factor has a last digit coprime to 10.


Given Data / Assumptions:

  • Each whole number’s last digit is assumed independent and uniformly distributed over {0,…,9}.
  • Allowed last digits (coprime to 10): {1, 3, 7, 9} (four choices).


Concept / Approach:
For the product to end with 1, 3, 7, or 9, every chosen last digit must be from {1, 3, 7, 9}. If any factor contributes a 0, 2, 4, 5, 6, or 8, the product gains a factor 2 and/or 5, making last digit even or 0/5.


Step-by-Step Solution:

Single number: P(last digit ∈ {1,3,7,9}) = 4/10 = 2/5Four independent choices ⇒ P = (2/5)^4 = 16/625


Verification / Alternative check:
Complement counts any occurrence of a digit with a factor 2 or 5; the direct approach above is simpler and exact.


Why Other Options Are Wrong:
Values like 4/25 or 8/125 correspond to fewer than four constraints; 1/210 is unrelated to base-10 residue counting.


Common Pitfalls:
Allowing last digit 5 or an even digit in any factor, which invalidates membership in {1,3,7,9} for the product.


Final Answer:
16/625

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