CAT 2024 Slot 2 · DILR · Question 14

Solving the set

The grid has Row 1: Columns 1–4; Row 2: Columns 2–4; Row 3: Columns 3–4; Row 4: Column 4 only. Write (r, c) for the slot in Row r, Column c.

Step 1 – Place 10. Every slot (r, c) is smaller than (r, 4) (rows increase to the right), and (r, 4) is smaller than (1, 4) (columns decrease downwards). So (1, 4) is the largest slot: 10 is at (1, 4).

Step 2 – Place 2 and 3. They are not in Row 1 or Column 4 (fact 4), so they lie in (2, 2), (2, 3) or (3, 3). Now (2, 3) is larger than both (2, 2) and (3, 3). If 3 were at (2, 3), the other two slots would need numbers below 3, i.e., 1 and 2, but 1 must be in Row 1 or Column 4 (fact 3). So 2 and 3 are at (2, 2) and (3, 3) in some order.

Step 3 – Place 9. Only (1, 3) and (2, 4) have no larger slot other than (1, 4), so 9 is at one of them.

• If 9 is at (2, 4): 7 and 8 cannot be in Row 2 or Column 4 (fact 5), so they must be in Row 1, Columns 1–3. The numbers left (1, 4, 5, 6) go into one free Row 1 slot, (2, 3), (3, 4) and (4, 4): no row has two free slots, so 4 and 6 cannot share a row (fact 6). Not possible.

• So 9 is at (1, 3). Then 7 and 8 are not in Row 1 or Column 3, so they are in Column 4, rows 2–4.

Step 4 – Place 1. The smallest number must be at the bottom of a column, i.e., (1, 1), (2, 2), (3, 3) or (4, 4); with fact 3 it is (1, 1) or (4, 4).

• If 1 is at (1, 1): 4 and 6 need a row with two free slots. Row 1 has only (1, 2) free, so they go to Row 2: (2, 3) = 4 and (2, 4) = 6. But 7 and 8 must then be below 6 in Column 4, which breaks the decreasing order. Not possible.

• So 1 is at (4, 4). Then 7 and 8 are at (3, 4) and (2, 4); decreasing order gives (2, 4) = 8, (3, 4) = 7.

Step 5 – The numbers 4, 5, 6 remain for (1, 1), (1, 2) and (2, 3). 4 and 6 are in the same row, so they are (1, 1) = 4 and (1, 2) = 6, and (2, 3) = 5. All conditions are satisfied.

	Column 1	Column 2	Column 3	Column 4
Row 1	4	6	9	10
Row 2	—	2 or 3	5	8
Row 3	—	—	3 or 2	7
Row 4	—	—	—	1

This question

Row sums from the final grid: Row 1 = 4 + 6 + 9 + 10 = 29; Row 2 = (2 or 3) + 5 + 8 = 15 or 16; Row 3 = (3 or 2) + 7 = 10 or 9; Row 4 = 1.

Row 4 has the least sum in every case.

Hence, the answer is 4.