Prove that if a1, a2,..., an are n ≥ 2 integers such that ai ≡ 1 (mod 3) for every integer i (1 ≤ i ≤ n), then a1a2 ··· an ≡ 1 (mod 3)

To prove this statement, we can use mathematical induction.

For n = 2, we have a1a2 ≡ 1 (mod 3) since both a1 and a2 are congruent to 1 modulo 3.

Assume the statement is true for n = k. That is, if a1, a2, ..., ak are integers such that ai ≡ 1 (mod 3) for every integer i (1 ≤ i ≤ k), then a1a2···ak ≡ 1 (mod 3).

Now, consider n = k + 1. We have ak+1 ≡ 1 (mod 3), since ak+1 is one of the given integers, and a1, a2, ..., ak are also congruent to 1 modulo 3. Thus,

a1a2···akak+1 ≡ (a1a2···ak)ak+1 ≡ 1×1 ≡ 1 (mod 3)

by the induction hypothesis.

Therefore, the statement is true for all n ≥ 2.

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