Problem:
A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is , where and are relatively prime positive integers. Find .
Solution:
We are asked to compute the probability that the product of four fair 6-sided dice rolls is a perfect square. Denote the set of outcomes as $ {1, 2, 3, 4, 5, 6} $. Let us analyze this problem through the lens of \textbf{exponents in prime factorization}.
A product is a perfect square if all exponents in its prime factorization are even.
The prime factorizations are:
Note: The number 5 introduces a new prime that cannot be canceled unless it appears an even number of times. Thus, we split the total number of outcomes into three cases based on the number of times 5 appears.
Case 1: 5 appears 4 times.
The only possible roll is . Since is a perfect square, this is valid.
Number of outcomes: .
Case 2: 5 appears exactly 2 times.
We choose 2 of the 4 dice to be 5, and the other 2 dice must produce a product that is a perfect square from the set .
We compute all valid combinations:
Total such outcomes: .
Number of ways to choose which 2 positions are 5: .
Total outcomes in this case: .
Case 3: No 5 appears.
We roll the 5-sided die four times. We want the product to be a perfect square.
Let be the number of perfect-square-producing sequences of length using only . We define the recurrence:
since for each perfect square product of rolls, appending a 1 or 4 keeps it a perfect square (2 options), and for each non-square product, there's a unique value (among 2,3,6) that corrects it to a square.
Starting from , we compute:
So there are outcomes in Case 3.
Sum across all cases to find the total number of favorable outcomes: .
Total possible outcomes: .
So the desired probability is .
The problems on this page are the property of the MAA's American Mathematics Competitions