Puzzle for February 21, 2023  ( )

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

eq.1) A + B + C + D + E + F = 26 eq.2) C = A + D + F eq.3) A - D = D - F eq.4) E - A = A + F eq.5) D + E = A + B + C eq.6) C + F = A + 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.6, replace C with A + D + F (from eq.2): A + D + F + F = A + D + E which becomes A + D + 2×F = A + D + E Subtract A and D from each side of the above equation: A + D + 2×F - A - D = A + D + E - A - D which simplifies to eq.6a) 2×F = E


  

Hint #2


In eq.4, replace E with 2×F: 2×F - A = A + F In the above equation, add A to both sides, and subtract F from both sides: 2×F - A + A - F = A + F + A - F which simplifies to F = 2×A


  

Hint #3


In eq.6a, substitute (2×A) for F: 2×(2×A) = E which makes 4×A = E


  

Hint #4


Substitute 2×A for F in eq.3: A - D = D - 2×A Add D and 2×A to both sides of the equation above: A - D + D + 2×A = D - 2×A + D + 2×A which becomes 3×A = 2×D Divide both sides by 2: 3×A ÷ 2 = 2×D ÷ 2 which makes 1½×A = D


  

Hint #5


Substitute 1½×A for D, and 2×A for F in eq.2: C = A + 1½×A + 2×A which makes C = 4½×A


  

Hint #6


Substitute 1½×A for D, 4×A for E, and 4½×A for C in eq.5: 1½×A + 4×A = A + B + 4½×A which becomes 5½×A = 5½×A + B Subtract 5½×A from each side of the above equation: 5½×A - 5½×A = 5½×A + B - 5½×A which makes 0 = B


  

Solution

Substitute 0 for B, 4½×A for C, 1½×A for D, 4×A for E, and 2×A for F in eq.1: A + 0 + 4½×A + 1½×A + 4×A + 2×A = 26 which simplifies to 13×A = 26 Divide both sides of the above equation by 13: 13×A ÷ 13 = 26 ÷ 13 which makes A = 2 making C = 4½×A = 4½ × 2 = 9 D = 1½×A = 1½ × 2 = 3 E = 4×A = 4 × 2 = 8 F = 2×A = 2 × 2 = 4 and ABCDEF = 209384