Puzzle for October 14, 2022  ( )

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

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

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

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Hint #1


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


  

Hint #2


In eq.4a, replace 2×F with C + B (from eq.3a): A + C + B = B + E Subtract B from each side of the equation above: A + C + B - B = B + E - B which becomes eq.4b) A + C = E


  

Hint #3


In eq.1, replace A + C with E (from eq.4b): D = E


  

Hint #4


In eq.2, substitute D for E: D - A = D - D which becomes D - A = 0 Add D to both sides of the above equation: D - A + D = 0 + D which makes A = D


  

Hint #5


Substitute D for A and E in eq.4b: D + C = D Subtract D from each side of the above equation: D + C - D = D - D which makes C = 0


  

Hint #6


Substitute D for E and A, and 0 for C in eq.6: D ÷ D = B + 0 - D which becomes 1 = B - D Add D to both sides of the equation above: 1 + D = B - D + D which becomes eq.6a) 1 + D = B


  

Hint #7


Substitute D for A and E, and 1 + D for B (from eq.6a) in eq.4a: D + 2×F = 1 + D + D which becomes D + 2×F = 1 + 2×D Subtract D from both sides of the above equation: D + 2×F - D = 1 + 2×D - D which becomes 2×F = 1 + D Divide both sides by 2: 2×F ÷ 2 = (1 + D) ÷ 2 which becomes eq.4c) F = (1 + D) ÷ 2 which is equivalent to eq.4d) F = ½ + ½×D


  

Hint #8


In eq.5, substitute (1 + D) for B (from eq.6a), ((1 + D) ÷ 2) for the 1st F (from eq.4c), 0 for C, D for E, and (½ + ½×D) for the 2nd F (from eq.4d): (1 + D) ÷ ((1 + D) ÷ 2) = 0 + D - (½ + ½×D) which becomes 2 = D - ½ - ½×D which becomes 2 = ½×D - ½ Add ½ to both sides of the equation above: 2 + ½ = ½×D - ½ + ½ which makes 2½ = ½×D Multiply both sides by 2: 2 × 2½ = 2 × ½×D which makes 5 = D


  

Solution

Since D = 5, then: A = E = D = 5 B = 1 + D = 1 + 5 = 6 (from eq.6a) F = (1 + D) ÷ 2 = (1 + 5) ÷ 2 = 6 ÷ 2 = 3 (from eq.4c) and ABCDEF = 560553