Probability of x^2y^2z^2 Being Divisible by 5: A Comprehensive Analysis

This article delves into the probability P(x^2y^2z^2 ≡ 0 mod 5), where x, y, and z belong to the set of numbers (123 5n),

Introduction to the Problem

We split the set (S {1, 2, , 5n}) into (5) disjoint subsets (S_0, S_1, S_2, S_3, S_4), based on their congruence modulo 5. Each subset corresponds to a specific remainder when divided by 5.

Modular Arithmetic: Deciphering Congruences

The set (S_i) is the collection of numbers in (S) that have the remainder (i) when divided by 5. Any number (n) can be written in one of the forms: (5k, 5k 1, 5k 2, 5k 3, 5k 4). When squared, the result depends on the form:

(5k 1 equiv 1 pmod{5}) and ((5k 1)^2 equiv 1 pmod{5}) (5k 2 equiv 2 pmod{5}) and ((5k 2)^2 equiv 4 pmod{5}) (5k 3 equiv 3 pmod{5}) and ((5k 3)^2 equiv 4 pmod{5}) (5k equiv 0 pmod{5}) and ((5k)^2 equiv 0 pmod{5}) (5k 4 equiv 4 pmod{5}) and ((5k 4)^2 equiv 1 pmod{5})

Probability Analysis

We need to determine the probability that the product (x^2y^2z^2) is divisible by 5. This requirement is met if any of the following conditions are satisfied:

(x^2 equiv 0 pmod{5}), (y^2 equiv 0 pmod{5}), (z^2 equiv 0 pmod{5}) (x^2 equiv 1 pmod{5}), (y^2 equiv 4 pmod{5}), (z^2 equiv 4 pmod{5}) (x^2 equiv 4 pmod{5}), (y^2 equiv 1 pmod{5}), (z^2 equiv 4 pmod{5}) (x^2 equiv 4 pmod{5}), (y^2 equiv 1 pmod{5}), (z^2 equiv 4 pmod{5})

Given that there are (n) numbers in each subset, the number of ways to choose one number from each subset is ({n choose 1}). The total number of ways to choose 3 numbers from the set is ({5n choose 3}).

The overall probability can be expressed as:

[P(x^2y^2z^2 equiv 0 pmod{5}) {n choose 1}{n choose 1}{n choose 1} / {5n choose 3} 6 cdot 2 cdot {2n choose 1}{2n choose 1}{n choose 1} / {5n choose 3} 25n^3 / {5n choose 3}]

Conclusion

This analysis effectively encapsulates the probability of x^2y^2z^2 being divisible by 5, given the constraints on x, y, and z.