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Difficulty: 5/102023 IOQM 2023 (Q27)

A quadruple (a, b, c, d) of distinct integers is said to be balanced if + = + d. Let S be any set of quadruples (a, b, c, d) where 1 < < < 20 and where the cardinality of S is 4411. Find the least number of balanced quadruples in S.

Guide / Hint

Hint 1: Start by analyzing the initial conditions and setting up the basic equations. a + = + a<b<d<c.

Hint 2: Look for algebraic properties, symmetry, or geometric theorems to simplify. a + = 37 (20, 17), (18, 19) → 2 → 1 way.

Hint 3: Proceed with the final algebraic steps to solve the system. a + solve for the final value (20, 16), (19, 17) → 2 → 1 way.

Solution

Step 1: a + = + a<b<d<c

Step 2: a + = 37 (20, 17), (18, 19) → 2 → 1 way

Step 3: a + = 36 (20, 16), (19, 17) → 2 → 1 way

Step 4: a + = 35 (20, 15), (19, 16), (18, 17) → 3 → 3C

Step 5: a + = 34 (20, 14), (19, 15), (18, 16) → 3 → 3C

Step 6: a + = 33 (20, 13), (19, 14)…(18, 15), (17, 16) → 4 → 4C

Step 7: a + = 32 (20, 12), (19, 13)…(17, 15) → 4

Step 8: a + = 31 (20, 11), (16, 15) → 5

Step 9: a + = 30 (20, 10), (16, 14) → 5

Step 10: a + = 29 → 6 (20, 9)… (15, 14) → 6

Step 11: a + = 28 → 6 (20, 8)\dots.. (15, 13) → 6

Step 12: a + = 27 → 7 (20, 7)\dots. (14, 13) → 7

Step 13: a + = 26 → 7

Step 14: a + = 25 → 8

Step 15: a + = 24 → 8

Step 16: a + = 23 → 9

Step 17: a + = 22 → 9

Step 18: a + = 21 → 10 (20, 1)…. (11, 10)

Step 19: a + = 20 → 9 (19, 1), (18, 2) \dots.. (11, 9)

Step 20: a + = 19 → 9 (18, 1)…. (11, 9)

Step 21: a + = 18 → 8 (17, 1), (10, 8)

Step 22: a+c=5→2 (4, 1) (3, 2)

Step 23: ( )

Step 24: Total balanced quadruple = 4 2 C2 + 3C2 \dots 9C2 + 10C2

Step 25: = 4 10C3 + 10C2

Step 26: 4 \times 10 \times 9 \times 8

Step 27: = + 45 = 525

Step 28: 4 – 4411 = 434

Step 29: For least balanced = 525 – 434 = 91

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