Puzzle for December 20, 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) B = C + F eq.3) F = C + E eq.4) D - E = F - D eq.5) B + C = D + E eq.6) D ÷ C = B ÷ A

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

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Hint #1


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


  

Hint #2


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


  

Hint #3


In eq.4, substitute 3×C for D: 3×C - E = F - 3×C Add 3×C to both sides of the equation above: 3×C - E + 3×C = F - 3×C + 3×C which becomes eq.4a) 6×C - E = F


  

Hint #4


Substitute 6×C - E for F (from eq.4a) in eq.3: 6×C - E = C + E In the above equation, add E to both sides, and subtract C from both sides: 6×C - E + E - C = C + E + E - C which makes 5×C = 2×E Divide both sides by 2: 5×C ÷ 2 = 2×E ÷ 2 which makes 2½×C = E


  

Hint #5


Substitute 2½×C for E in eq.4a: 6×C - 2½×C = F which makes 3½×C = F


  

Hint #6


Substitute 2½×C for E in eq.2a: B = 2×C + 2½×C which makes B = 4½×C


  

Hint #7


Substitute 3×C for D, and 4½×C for B in eq.6: 3×C ÷ C = 4½×C ÷ A which becomes 3 = 4½×C ÷ A Multiply both sides of the above equation by A: 3 × A = (4½×C ÷ A) × A which makes 3×A = 4½×C Divide both sides by 3: 3×A ÷ 3 = 4½×C ÷ 3 which makes A = 1½×C


  

Solution

Substitute 1½×C for A, 4½×C for B, 3×C for D, 2½×C for E, and 3½×C for F in eq.1: 1½×C + 4½×C + C + 3×C + 2½×C + 3½×C = 32 which simplifies to 16×C = 32 Divide both sides of the above equation by 16: 16×C ÷ 16 = 32 ÷ 16 which means C = 2 making A = 1½×C = 1½ × 2 = 3 B = 4½×C = 4½ × 2 = 9 D = 3×C = 3 × 2 = 6 E = 2½×C = 2½ × 2 = 5 F = 3½×C = 3½ × 2 = 7 and ABCDEF = 392657