Puzzle for November 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 = 34 eq.2) B = A + E eq.3) B + F = C + D eq.4) A + C = B + D + E eq.5) C - D + F = A + D - E eq.6) A = average (B, E, F)

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

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Hint #1


eq.6 may be written as: A = (B + E + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × A = 3 × (B + E + F) ÷ 3 which becomes eq.6a) 3×A = B + E + F


  

Hint #2


In eq.6a, replace B with A + E (from eq.2): 3×A = A + E + E + F which becomes 3×A = A + 2×E + F Subtract A from each side of the equation above: 3×A - A = A + 2×E + F - A which becomes 2×A = 2×E + F Divide both sides by 2: 2×A ÷ 2 = (2×E + F) ÷ 2 which becomes eq.6b) A = E + ½×F


  

Hint #3


In eq.2, replace A with E + ½×F (from eq.6b): B = E + ½×F + E which becomes eq.2a) B = 2×E + ½×F


  

Hint #4


In eq.4, substitute A + E for B (from eq.2): A + C = A + E + D + E which becomes A + C = A + 2×E + D Subtract A from both sides of the equation above: A + C - A = A + 2×E + D - A which becomes eq.4a) C = 2×E + D


  

Hint #5


Substitute 2×E + D for C (from eq.4a), and E + ½×F for A (from eq.6b) in eq.5: 2×E + D - D + F = E + ½×F + D - E which becomes 2×E + F = ½×F + D Subtract ½×F from both sides of the equation above: 2×E + F - ½×F = ½×F + D - ½×F which becomes eq.5a) 2×E + ½×F = D


  

Hint #6


Substitute 2×E + ½×F for D (from eq.5a) in eq.4a: C = 2×E + 2×E + ½×F which becomes eq.4b) C = 4×E + ½×F


  

Hint #7


Substitute 2×E + ½×F for B (from eq.2a), 4×E + ½×F for C (from eq.4b), and 2×E + ½×F for D (from eq.5a) in eq.3: 2×E + ½×F + F = 4×E + ½×F + 2×E + ½×F which becomes 2×E + 1½×F = 6×E + F Subtract 2×E and F from both sides of the equation above: 2×E + 1½×F - 2×E - F = 6×E + F - 2×E - F which makes ½×F = 4×E Multiply both sides by 2: 2 × ½×F = 2 × 4×E which makes F = 8×E


  

Hint #8


Substitute (8×E) for F in eq.6b: A = E + ½×(8×E) which becomes A = E + 4×E which makes A = 5×E


  

Hint #9


Substitute (8×E) for F in eq.2a: B = 2×E + ½×(8×E) which becomes B = 2×E + 4×E which makes B = 6×E


  

Hint #10


Substitute (8×E) for F in eq.4b: C = 4×E + ½×(8×E) which becomes C = 4×E + 4×E which makes C = 8×E


  

Hint #11


Substitute (8×E) for F in eq.5a: 2×E + ½×(8×E) = D which becomes 2×E + 4×E = D which makes 6×E = D


  

Solution

Substitute 5×E for A, 6×E for B and D, and 8×E for C and F in eq.1: 5×E + 6×E + 8×E + 6×E + E + 8×E = 34 which simplifies to 34×E = 34 Divide both sides of the above equation by 34: 34×E ÷ 34 = 34 ÷ 34 which means E = 1 making A = 5×E = 5 × 1 = 5 B = D = 6×E = 6 × 1 = 6 C = F = 8×E = 8 × 1 = 8 and ABCDEF = 568618