Clue 9: (C1, D1) sum = 7 → possible (2,5),(3,4),(4,3),(5,2).
Clue 2: A2 and A3 same symbol. So they could both be ★ or both non-★.
Clue 10: |B3-B4|=3.
She checks the original text: Clue 6 actually says: (E1, E2): Same number. That’s impossible under standard rules. So either it’s a trick — meaning E1 and E2 are the same number, so the row has a duplicate, meaning the “each row has 1..5 once” rule is for numbers? Or the puzzle uses numbers 1-5 with repetition allowed? But that breaks Latin square. Elites Grid LRDI 2023 Matrix Arrangement lesson...
Clue 1: (A1, A2) sum to 6. Possible pairs: (1,5), (2,4), (3,3), (4,2), (5,1). But clue 2 says A2 and A3 share the same symbol. Not yet a number lock.
That fixes it. Now E1 and E2 share a symbol, say S_E. E4 and E5 differ by 2 in number.
Riya slams the table. “Ah! That’s the trap. Clue 6 says ‘same number’ but that violates the row uniqueness. So either the puzzle allows duplicates (rare) or ‘same number’ means they are equal but then the row must have a duplicate — impossible. Therefore, clue 6 must be interpreted as ‘same symbol’, not same number!” Clue 9: (C1, D1) sum = 7 →
“The trick is to treat numbers and symbols as two interlocking Latin squares. Start with the most restrictive clue — here, the ★ per row/col plus product odd and sum clues. Use a 5x5 possibilities table. Never assume without checking row-column uniqueness for both attributes simultaneously.”
After 20 minutes of elimination (details omitted for brevity, but in a real LRDI, you’d use a 5x5 table and test constraints), the unique solution emerges:
Clue 6: (E1, E2) same number. So E1 = E2 = x. But rows must have 1..5 each exactly once. So x can be 1..5, but that means E3, E4, E5 are the other four numbers. Clue 10: |B3-B4|=3
Clue 7: (E4, E5) difference 2 → possible pairs: (1,3),(2,4),(3,1),(3,5),(4,2),(5,3).
Clue 3: B2<C2.