Puzzle for June 4, 2023  ( )

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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) B + E = A + C eq.3) E + F = B - F eq.4) C + D = A - D + E eq.5) A + D + F = B + C - A - F eq.6)* DE - F = A + B + C + D + F

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  DE is a 2-digit number (not D×E).

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Hint #1


Subtract E and C from each side of eq.2: B + E - E - C = A + C - E - C which becomes eq.2a) B - C = A - E   Add D to both sides of eq.4: C + D + D = A - D + E + D which becomes eq.4a) C + 2×D = A + E


  

Hint #2


Add the left and right sides of eq.2a to the left and right sides of eq.4a, respectively: C + 2×D + (B - C) = A + E + (A - E) which becomes eq.4b) 2×D + B = 2×A


  

Hint #3


Add F to both sides of eq.3: E + F + F = B - F + F which becomes eq.3a) E + 2×F = B   Subtract 2×D and A from both sides of eq.4a: C + 2×D - 2×D - A = A + E - 2×D - A which becomes eq.4c) C - A = E - 2×D


  

Hint #4


In eq.5, replace C - A with E - 2×D (from eq.4c): A + D + F = B + E - 2×D - F Add 2×D and F to both sides of the above equation: A + D + F + 2×D + F = B + E - 2×D - F + 2×D + F which becomes eq.5a) A + 3×D + 2×F = B + E


  

Hint #5


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


  

Hint #6


In eq.4b, substitute (2×E - 3×D) for A (from eq.5b): 2×D + B = 2×(2×E - 3×D) which becomes 2×D + B = 4×E - 6×D Subtract 2×D from both sides of the equation above: 2×D + B - 2×D = 4×E - 6×D - 2×D which becomes eq.4d) B = 4×E - 8×D


  

Hint #7


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


  

Hint #8


Substitute 4×E - 8×D for B (from eq.4d) in eq.3a: E + 2×F = 4×E - 8×D Subtract E from each side of the above equation: E + 2×F - E = 4×E - 8×D - E which becomes eq.3b) 2×F = 3×E - 8×D


  

Hint #9


eq.6 may be written as: 10×D + E - F = A + B + C + D + F In the above equation, add F to both sides, and subtract D from both sides: 10×D + E - F + F - D = A + B + C + D + F + F - D which becomes eq.6a) 9×D + E = A + B + C + 2×F


  

Hint #10


In eq.6a, substitute 2×E - 3×D for A (from eq.5b), 4×E - 8×D for B (from eq.4d), 3×E - 5×D for C (from eq.4e), and 3×E - 8×D for 2×F (from eq.3b): 9×D + E = 2×E - 3×D + 4×E - 8×D + 3×E - 5×D + 3×E - 8×D which simplifies to 9×D + E = 12×E - 24×D In the above equation, subtract E from both sides, and add 24×D to both sides: 9×D + E - E + 24×D = 12×E - 24×D - E + 24×D which makes 33×D = 11×E Divide both sides by 11: 33×D ÷ 11 = 11×E ÷ 11 which makes 3×D = E


  

Hint #11


Substitute (3×D) for E in eq.5b: A = 2×(3×D) - 3×D which becomes A = 6×D - 3×D which makes A = 3×D


  

Hint #12


Substitute (3×D) for E in eq.4d: B = 4×(3×D) - 8×D which becomes B = 12×D - 8×D which makes B = 4×D


  

Hint #13


Substitute (3×D) for E in eq.4e: C = 3×(3×D) - 5×D which becomes C = 9×D - 5×D which makes C = 4×D


  

Hint #14


Substitute (3×D) for E in eq.3b: 2×F = 3×(3×D) - 8×D which becomes 2×F = 9×D - 8×D which makes 2×F = D Divide both sides of the above equation by 2: 2×F ÷ 2 = D ÷ 2 which makes F = ½×D


  

Solution

Substitute 3×D for A and E, 4×D for B and C, and ½×D for F in eq.1: 3×D + 4×D + 4×D + D + 3×D + ½×D = 31 which simplifies to 15½×D = 31 Divide both sides of the above equation by 15½: 15½×D ÷ 15½ = 31 ÷ 15½ which means D = 2 making A = E = 3×D = 3 × 2 = 6 B = C = 4×D = 4 × 2 = 8 F = ½×D = ½ × 2 = 1 and ABCDEF = 688261