The total number of three-digit numbers, divisible by 3, which can be formed using the digits 1, 3, 5, 8, if repetition of digits is allowed, is

We need 3-digit numbers using {1, 3, 5, 8} with repetition, divisible by 3.

Digits mod 3: 1→1, 3→0, 5→2, 8→2. Residue counts: 0→1 choice, 1→1 choice, 2→2 choices.

Valid residue triplets summing to 0 mod 3:

(0,0,0): 1×1×1 = 1

(0,1,2): 6 arrangements × 1×1×2 = 12

(1,1,1): 1×1×1 = 1

(2,2,2): 2×2×2 = 8

Total = 1 + 12 + 1 + 8 = 22

The answer is $$\mathbf{22}$$.

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