Puzzle for December 19, 2023  ( )

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

eq.1) A + B + C + D + E + F = 32 eq.2) F = C + E eq.3) B = E + F eq.4) A = B + C eq.5) D = A + C eq.6) C + D = B + E

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 F with C + E (from eq.2): B = E + C + E which becomes eq.3a) B = 2×E + C


  

Hint #2


In eq.4, replace B with 2×E + C (from eq.3a): A = 2×E + C + C which becomes eq.4a) A = 2×E + 2×C


  

Hint #3


In eq.5, substitute 2×E + 2×C for A (from eq.4a): D = 2×E + 2×C + C which becomes eq.5a) D = 2×E + 3×C


  

Hint #4


Substitute 2×E + 3×C for D (from eq.5a), and 2×E + C for B (from eq.3a) in eq.6: C + 2×E + 3×C = 2×E + C + E which becomes 4×C + 2×E = 3×E + C Subtract 2×E and C from each side of the equation above: 4×C + 2×E - 2×E - C = 3×E + C - 2×E - C which makes 3×C = E


  

Hint #5


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


  

Hint #6


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


  

Hint #7


Substitute (3×C) for E in eq.4a: A = 2×(3×C) + 2×C which becomes A = 6×C + 2×C which makes A = 8×C


  

Hint #8


Substitute (3×C) for E in eq.5a: D = 2×(3×C) + 3×C which becomes D = 6×C + 3×C which makes D = 9×C


  

Solution

Substitute 8×C for A, 7×C for B, 9×C for D, 3×C for E, and 4×C for F in eq.1: 8×C + 7×C + C + 9×C + 3×C + 4×C = 32 which simplifies to 32×C = 32 Divide both sides of the above equation by 32: 32×C ÷ 32 = 32 ÷ 32 which means C = 1 making A = 8×C = 8 × 1 = 8 B = 7×C = 7 × 1 = 7 D = 9×C = 9 × 1 = 9 E = 3×C = 3 × 1 = 3 F = 4×C = 4 × 1 = 4 and ABCDEF = 871934