Puzzle for March 22, 2023  ( )

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

eq.1) A + B + C + D + E + F = 22 eq.2) B = C + F eq.3) E = D + F eq.4) D - B = B - C eq.5) C - F = D - C eq.6) B + D = A + E

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

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Hint #1


In eq.6, replace E with D + F (from eq.3): B + D = A + D + F Subtract D from each side of the equation above: B + D - D = A + D + F - D which becomes eq.6a) B = A + F


  

Hint #2


In eq.2, replace B with A + F (from eq.6a): A + F = C + F Subtract F from each side of the above equation: A + F - F = C + F - F which becomes A = C


  

Hint #3


Add C to both sides of eq.5: C - F + C = D - C + C which becomes 2×C - F = D In eq.3, substitute 2×C - F for D: E = 2×C - F + F which becomes E = 2×C


  

Hint #4


Substitute C for A, and 2×C for E in eq.6: B + D = C + 2×C which becomes B + D = 3×C Subtract B from both sides of the above equation: B + D - B = 3×C - B which becomes eq.6b) D = 3×C - B


  

Hint #5


Substitute 3×C - B for D (from eq.6b) in eq.4: 3×C - B - B = B - C which becomes 3×C - 2×B = B - C Add 2×B and C to both sides of the above equation: 3×C - 2×B + 2×B + C = B - C + 2×B + C which becomes 4×C = 3×B Divide both sides by 3: 4×C ÷ 3 = 3×B ÷ 3 which makes 1⅓×C = B


  

Hint #6


Substitute 1⅓×C for B in eq.6b: D = 3×C - 1⅓×C which makes D = 1⅔×C


  

Hint #7


Substitute 1⅓×C for B in eq.2: 1⅓×C = C + F Subtract C from each side of the above equation: 1⅓×C - C = C + F - C which makes ⅓×C = F


  

Solution

Substitute C for A, 1⅓×C for B, 1⅔×C for D, 2×C for E, and ⅓×C for F in eq.1: C + 1⅓×C + C + 1⅔×C + 2×C + ⅓×C = 22 which simplifies to 7⅓×C = 22 Divide both sides of the above equation by 7⅓: 7⅓×C ÷ 7⅓ = 22 ÷ 7⅓ which means C = 3 making A = C = 3 B = 1⅓×C = 1⅓ × 3 = 4 D = 1⅔×C = 1⅔ × 3 = 5 E = 2×C = 2 × 3 = 6 F = ⅓×C = ⅓ × 3 = 1 and ABCDEF = 343561