Puzzle for November 23, 2022  ( )

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

eq.1) A + B + C + D + E + F = 31 eq.2) E + F = B + C eq.3) D + E = A + F eq.4) C + D = B + E eq.5) B + F = A + D + E eq.6) A + B = C + D + 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 C + D with B + E (from eq.4): A + B = B + E + E which becomes A + B = B + 2×E Subtract B from each side of the above equation: A + B - B = B + 2×E - B which makes eq.6a) A = 2×E


  

Hint #2


In eq.5, replace D + E with A + F (from eq.3): B + F = A + A + F which becomes B + F = 2×A + F Subtract F from each side of the equation above: B + F - F = 2×A + F - F which makes eq.5a) B = 2×A


  

Hint #3


In eq.5a, substitute (2×E) for A (from eq.6a): B = 2×(2×E) which makes eq.5b) B = 4×E


  

Hint #4


Substitute 4×E for B (from eq.5b) in eq.2: E + F = 4×E + C Subtract E from both sides of the above equation: E + F - E = 4×E + C - E which becomes eq.2a) F = 3×E + C


  

Hint #5


Substitute 4×E for B (from eq.5b) in eq.4: C + D = 4×E + E which becomes C + D = 5×E Subtract C from both sides of the equation above: C + D - C = 5×E - C which becomes eq.4a) D = 5×E - C


  

Hint #6


Substitute 5×E - C for D (from eq.4a), 2×E for A (from eq.6a), and 3×E + C for F (from eq.2a) in eq.3: 5×E - C + E = 2×E + 3×E + C which becomes 6×E - C = 5×E + C In the above equation, add C to both sides, and subtract 5×E from both sides: 6×E - C + C - 5×E = 5×E + C + C - 5×E which simplifies to E = 2×C


  

Hint #7


Substitute (2×C) for E in eq.6a: A = 2×(2×C) which makes A = 4×C


  

Hint #8


Substitute (2×C) for E in eq.5b: B = 4×(2×C) which makes B = 8×C


  

Hint #9


Substitute (2×C) for E in eq.4a: D = 5×(2×C) - C which becomes D = 10×C - C which makes D = 9×C


  

Hint #10


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


  

Solution

Substitute 4×C for A, 8×C for B, 9×C for D, 2×C for E, and 7×C for F in eq.1: 4×C + 8×C + C + 9×C + 2×C + 7×C = 31 which simplifies to 31×C = 31 Divide both sides of the above equation by 31: 31×C ÷ 31 = 31 ÷ 31 which means C = 1 making A = 4×C = 4 × 1 = 4 B = 8×C = 8 × 1 = 8 D = 9×C = 9 × 1 = 9 E = 2×C = 2 × 1 = 2 F = 7×C = 7 × 1 = 7 and ABCDEF = 481927