Puzzle for September 3, 2021  ( )

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

eq.1) A + B + C + D + E + F = 39 eq.2) B + E = A eq.3) F = A + E eq.4) C + D = A + F eq.5) A + D = B + C + F – A eq.6) B = average (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.5, replace F with A + E (from eq.3): A + D = B + C + A + E – A which becomes A + D = B + C + E which may be written as eq.5a) A + D = B + E + C


  

Hint #2


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


  

Hint #3


In eq.3, substitute B + E for A (from eq.2): F = B + E + E which becomes eq.3a) F = B + 2×E


  

Hint #4


In eq.4, substitute D for C, B + E for A (from eq.2), and B + 2×E for F (from eq.3a): D + D = B + E + B + 2×E which becomes eq.4a) 2×D = 2×B + 3×E


  

Hint #5


eq.6 may be written as: B = (D + E) ÷ 2 Multiply both sides of the above equation by 2: 2 × B = 2 × (D + E) ÷ 2 which becomes eq.6a) 2×B = D + E


  

Hint #6


Substitute D + E for 2×B (from eq.6a) into eq.4a: 2×D = D + E + 3×E which becomes 2×D = D + 4×E Subtract D from each side of the above equation: 2×D – D = D + 4×E – D which makes D = 4×E and also makes C = D = 4×E


  

Hint #7


Substitute 4×E for D in eq.6a: 2×B = 4×E + E which becomes 2×B = 5×E Divide both sides of the above equation by 2: 2×B ÷ 2 = 5×E ÷ 2 which makes B = 2½×E


  

Hint #8


Substitute 2½×E for B in eq.3a: F = 2½×E + 2×E which makes F = 4½×E


  

Hint #9


Substitute 2½×E for B in eq.2: 2½×E + E = A which makes 3½×E = A


  

Solution

Substitute 3½×E for A, 2½×E for B, 4×E for C and D, and 4½×E for F in eq.1: 3½×E + 2½×E + 4×E + 4×E + E + 4½×E = 39 which simplifies to 19½×E = 39 Divide both sides of the above equation by 19½: 19½×E ÷ 19½ = 39 ÷ 19½ which means E = 2 making A = 3½×E = 3½ × 2 = 7 B = 2½×E = 2½ × 2 = 5 C = D = 4×E = 4 × 2 = 8 F = 4½×E = 4½ × 2 = 9 and ABCDEF = 758829