Puzzle for January 31, 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 + D eq.3) E = A + D eq.4) B = A + E eq.5) F - A = A + D eq.6) D + E = A + C - D

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 A + D (from eq.3): D + A + D = A + C - D which becomes 2×D + A = A + C - D In the above equation, subtract A from both sides, and add D to both sides: 2×D + A - A + D = A + C - D - A + D which simplifies to 3×D = C


  

Hint #2


In eq.2, replace C with 3×D: B = 3×D + D which makes B = 4×D


  

Hint #3


In eq.4, substitute 4×D for B, and A + D for E (from eq.3): 4×D = A + A + D which becomes 4×D = 2×A + D Subtract D from each side of the equation above: 4×D - D = 2×A + D - D which becomes 3×D = 2×A Divide both sides by 2: 3×D ÷ 2 = 2×A ÷ 2 which makes 1½×D = A


  

Hint #4


Substitute 1½×D for A in eq.3: E = 1½×D + D which makes E = 2½×D


  

Hint #5


Substitute 1½×D for A in eq.5: F - 1½×D = 1½×D + D which becomes F - 1½×D = 2½×D Add 1½×D to both sides of the above equation: F - 1½×D + 1½×D = 2½×D + 1½×D which makes F = 4×D


  

Solution

Substitute 1½×D for A, 4×D for B and F, 3×D for C, and 2½×D for E in eq.1: 1½×D + 4×D + 3×D + D + 2½×D + 4×D = 32 which simplifies to 16×D = 32 Divide both sides of the above equation by 16: 16×D ÷ 16 = 32 ÷ 16 which means D = 2 making A = 1½×D = 1½ × 2 = 3 B = F = 4×D = 4 × 2 = 8 C = 3×D = 3 × 2 = 6 E = 2½×D = 2½ × 2 = 5 and ABCDEF = 386258