Puzzle for December 29, 2022  ( )

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

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

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 B with A + F (from eq.1): A + C = A + F + E Subtract A from each side of the equation above: A + C - A = A + F + E - A which becomes eq.2a) C = F + E


  

Hint #2


In eq.3, replace C with F + E (from eq.2a): E + F = F + E - F which becomes E + F = E Subtract E from both sides of the equation above: E + F - E = E - E which makes F = 0


  

Hint #3


In eq.1, substitute 0 for F: B = A + 0 which means B = A


  

Hint #4


In eq.2a, substitute 0 for F: C = 0 + E which means C = E


  

Hint #5


In eq.4, substitute 0 for F: C + 0 = B + D - 0 which becomes eq.4a) C = B + D


  

Hint #6


eq.6 may be written as: A - F = (B + D + C) ÷ E Substitute 0 for F, C for B + D (from eq.4a), and C for E in eq.6: A - 0 = (C + C) ÷ C which becomes A = (2×C) ÷ C which makes A = 2 and also makes B = A = 2


  

Hint #7


Substitute 2 for B in eq.4a: C = 2 + D and also makes eq.4b) E = C = 2 + D


  

Hint #8


eq.5 may be written as: D = (B + C + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × D = 3 × (B + C + E) ÷ 3 which becomes eq.5a) 3×D = B + C + E


  

Hint #9


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


  

Solution

Substitute 6 for D in eq.4b: E = C = 2 + 6 which makes E = C = 8 and makes ABCDEF = 228680