Correct answer: B. (5)
Explanation: The direct answer is option B: 5 pairs. We need ordered pairs \((x,y)\) with \(x,y\in\{1,2,3,4\}\) and \(x+y\) divisible by 3. Possible sums are 3, 6, or 9, but with values from 1 to 4, only 3 and 6 occur. Sum 3 gives \((1,2),(2,1)\). Sum 6 gives \((2,4),(3,3),(4,2)\). Thus the total is \(2+3=5\). Option A, 4, misses one valid pair. Option B, 5, lists all valid ordered pairs and is correct. Option C, 6, and option D, 7, include extra pairs whose sums are not divisible by 3. Order matters: \((1,2)\) and \((2,1)\) are different pairs. A useful exam cue is to group numbers by their remainders modulo 3, but direct checking here is quickest.