Sixteen or Twenty-Two: Adjudicating a Number Puzzle in Two Minutes
Every Thursday the ninth-grade math class of 24 works six number puzzles in ten minutes - about 100 seconds each - and then defends answers. Puzzle 4 is a three-by-three grid whose rows read 6, 3, 18 / 5, 4, 20 / 8, 2, ? . Ana answers 16: in each row the first number times the second gives the third (6 × 3 = 18, 5 × 4 = 20, so 8 × 2 = 16). Ben answers 22: the third column climbs 18, 20, so it keeps climbing by 2, and the rest of the grid is 'just there to distract you.' The class splits 15 to 9 in Ana's favour, and Ben's nine supporters are the louder group. The teacher has under two minutes to settle it and wants the settlement to be a procedure rather than an appeal to authority, because a new puzzle will be defended next week.
1. Adjudicate. Which answer is better supported, and what principle decides between two rules that both fit the printed numbers?
Read the worked answer
16 is better supported, and the deciding principle is how much of the printed information each rule uses.
Ana's rule - first × second = third - explains 6 × 3 = 18 and 5 × 4 = 20, then predicts 16 for the third row. It leaves no number in the grid unexplained. The whole grid is evidence for it, and it makes a definite, checkable prediction that the third row could have contradicted.
Ben's rule uses two numbers, 18 and 20, and asserts a constant step of 2. Any three numbers can be fitted by a constant step, so his 'pattern' was guaranteed to exist whatever the grid contained - it is a line drawn through two points, not evidence. His rule also declares six of the eight printed numbers irrelevant, which is the visible tell: a rule that must ignore most of the data to survive is weaker than one that uses it.
The asymmetry is the whole argument. Ana's rule could have been wrong; Ben's could not fail.
2. Give the class a 15-second tiebreak they can run during the test.
Read the worked answer
Two questions, then a direction check.
1. How many printed numbers does my rule ignore? If an entire row or column is unused, downgrade the rule. Ana's rule touches all three rows; Ben's touches one column and dismisses six numbers as scenery.
2. Could this rule have been wrong? A rule that fits any grid - a constant step fitted to two numbers, or 'the rest is a distraction' - predicts nothing and cannot be evidence. Prefer the rule that a contradicting third row would have killed.
Direction check. Apply the surviving rule down the columns as well as across the rows. Ana's rule does not produce a consistent column pattern, and that is fine: one well-supported relation is enough, and the absence of a second relation does not weaken the first. What would weaken it is a number inside the rows that contradicted it - there is none.
Edge case worth naming: if both candidate rules use the whole grid and give different answers, the item is flawed. Say so rather than guessing, because the habit of forcing a single answer trains the wrong reflex.
3. Next week's puzzle has rows 4, 3, 7 / 9, 5, 14 / 6, 8, ? . A student insists the answer is 28 because the third column doubles (7, 14, 28). Use the tiebreak to resolve it and say what the item is really testing.
Read the worked answer
Test the operation first. 4 + 3 = 7 and 9 + 5 = 14 both hold, so the row rule is addition, not multiplication, and it predicts 6 + 8 = 14. It uses every printed number and is applied twice before the prediction is made.
The 28 answer treats the third column as doubling. That rule rests on two printed numbers (7 and 14) and ignores the other six, and two numbers are equally consistent with +7, ×2, or a next term of 21 - there is no evidence to choose among them. It also cannot explain why 7 and 14 are there at all, since under the row rule both are produced by the first two entries of their rows.
What the item tests is rule selection under a planted distractor, not arithmetic: the third column was built to look patterned, and it will look even more patterned once a student has chosen to see doubling in it. The transferable habit is to fit the operation on the first two rows, confirm it on the third, and only then read off the missing cell.