Puzzle for June 27, 2023  ( )

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Find the 6-digit number ABCDEF by solving the following equations:

eq.1) A + B + C + D + E + F = 29 eq.2) A + B = C eq.3) C - D = A - B eq.4) D - A = F - D eq.5) F - A = A - B eq.6) E + F = C + D - F

A, B, C, D, E, and F each represent a one-digit non-negative integer.

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Hint #1


In eq.3, replace C with A + B (from eq.2): A + B - D = A - B Subtract A from both sides of the above equation: A + B - D - A = A - B - A which becomes B - D = -B Add D and B to both sides: B - D + D + B = -B + D + B which makes 2×B = D


  

Hint #2


In eq.4, replace D with 2×B: 2×B - A = F - 2×B Add 2×B and A to both sides of the above equation: 2×B - A + 2×B + A = F - 2×B + 2×B + A which becomes 4×B = F + A Subtract F from both sides: 4×B - F = F + A - F which becomes eq.4a) 4×B - F = A


  

Hint #3


In eq.5, substitute (4×B - F) for A (from eq.4a): F - (4×B - F) = (4×B - F) - B which becomes F - 4×B + F = 4×B - F - B which becomes 2×F - 4×B = 3×B - F Add 4×B and F to both sides of the equation above: 2×F - 4×B + 4×B + F = 3×B - F + 4×B + F which makes 3×F = 7×B Divide both sides by 3: 3×F ÷ 3 = 7×B ÷ 3 which makes F = 2⅓×B


  

Hint #4


Substitute 2⅓×B for F in eq.4a: 4×B - 2⅓×B = A which makes 1⅔×B = A


  

Hint #5


Substitute 1⅔×B for A in eq.2: 1⅔×B + B = C which makes 2⅔×B = C


  

Hint #6


Substitute 2⅓×B for F, 2⅔×B for C, and 2×B for D in eq.6: E + 2⅓×B = 2⅔×B + 2×B - 2⅓×B which becomes E + 2⅓×B = 2⅓×B Subtract 2⅓×B from each side of the equation above: E + 2⅓×B - 2⅓×B = 2⅓×B - 2⅓×B which makes E = 0


  

Solution

Substitute 1⅔×B for A, 2⅔×B for C, 2×B for D, 0 for E, and 2⅓×B for F in eq.1: 1⅔×B + B + 2⅔×B + 2×B + 0 + 2⅓×B = 29 which simplifies to 9⅔×B = 29 Divide both sides of the above equation by 9⅔: 9⅔×B ÷ 9⅔ = 29 ÷ 9⅔ which means B = 3 making A = 1⅔×B = 1⅔ × 3 = 5 C = 2⅔×B = 2⅔ × 3 = 8 D = 2×B = 2 × 3 = 6 F = 2⅓×B = 2⅓ × 3 = 7 and ABCDEF = 538607