Puzzle for September 4, 2020  ( )

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

eq.1) A + B + C + D + E + F = 23 eq.2) B + E = A + D eq.3) F – C = A + C eq.4) C + D = B – C eq.5) D + E = A + C + F eq.6) B – A – C = A + F

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

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Hint #1


Add C to both sides of eq.3: F – C + C = A + C + C which becomes eq.3a) F = A + 2×C   In eq.6, replace F with A + 2×C (from eq.3a): B – A – C = A + A + 2×C which becomes B – A – C = 2×A + 2×C Add A and C to both sides of the above equation: B – A – C + A + C = 2×A + 2×C + A + C which becomes eq.6a) B = 3×A + 3×C


  

Hint #2


In eq.4, replace B with 3×A + 3×C (from eq.6a): C + D = 3×A + 3×C – C which becomes C + D = 3×A + 2×C Subtract C from each side of the equation above: C + D – C = 3×A + 2×C – C which becomes eq.4a) D = 3×A + C


  

Hint #3


In eq.2, substitute 3×A + 3×C for B (from eq.6a), and 3×A + C for D (from eq.4a): 3×A + 3×C + E = A + 3×A + C which becomes 3×A + 3×C + E = 4×A + C Subtract 3×A and 3×C from each side of the above equation: 3×A + 3×C + E – 3×A – 3×C = 4×A + C – 3×A – 3×C which becomes eq.2a) E = A – 2×C


  

Hint #4


Substitute 3×A + C for D (from eq.4a), A – 2×C for E (from eq.2a), and A + 2×C for F (from eq.3a) in eq.5: 3×A + C + A – 2×C = A + C + A + 2×C which becomes 4×A – C = 2×A + 3×C In the above equation, add C to both sides, and subtract 2×A from both sides: 4×A – C + C – 2×A = 2×A + 3×C + C – 2×A which makes 2×A = 4×C Divide both sides by 2: 2×A ÷ 2 = 4×C ÷ 2 which makes A = 2×C


  

Hint #5


Substitute (2×C) for A in eq.6a: B = 3×(2×C) + 3×C which becomes B = 6×C + 3×C which makes B = 9×C


  

Hint #6


Substitute (2×C) for A in eq.4a: D = 3×(2×C) + C which becomes D = 6×C + C which makes D = 7×C


  

Hint #7


Substitute 2×C for A in eq.2a: E = 2×C – 2×C which makes E = 0


  

Hint #8


Substitute 2×C for A in eq.3a: F = 2×C + 2×C which makes F = 4×C


  

Solution

Substitute 2×C for A, 9×C for B, 7×C for D, 0 for E, 4×C for F in eq.1: 2×C + 9×C + C + 7×C + 0 + 4×C = 23 which simplifies to 23×C = 23 Divide both sides of the above equation by 23: 23×C ÷ 23 = 23 ÷ 23 which means C = 1 making A = 2×C = 2 × 1 = 2 B = 9×C = 9 × 1 = 9 D = 7×C = 7 × 1 = 7 F = 4×C = 4 × 1 = 4 and ABCDEF = 291704