Puzzle for January 5, 2023  ( )

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

eq.1) A + B + C + D + E + F = 17 eq.2) D = A + C + F eq.3) A + F = B + C eq.4) B - C = A + E - F eq.5) A - B + D = B + C + F eq.6) D + F - B = A + B - 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 D with A + C + F (from eq.2): A - B + A + C + F = B + C + F which becomes 2×A - B + C + F = B + C + F In the above equation, add B to both sides, and subtract C and F from both sides: 2×A - B + C + F + B - C - F = B + C + F + B - C - F which simplifies to 2×A = 2×B Divide both sides by 2: 2×A ÷ 2 = 2×B ÷ 2 which makes A = B


  

Hint #2


In eq.3, replace B with A: A + F = A + C Subtract A from each side of the equation above: A + F - A = A + C - A which makes F = C


  

Hint #3


In eq.4, substitute A for B, and C for F: A - C = A + E - C In the above equation, subtract A from both sides, and add C to both sides: A - C - A + C = A + E - C - A + C which simplifies to 0 = E


  

Hint #4


Substitute C for F in eq.2: D = A + C + C which becomes eq.2a) D = A + 2×C


  

Hint #5


Substitute C for F, A for B, and 0 for E in eq.6: D + C - A = A + A - 0 which becomes D + C - A = 2×A Add A to both sides of the above equation: D + C - A + A = 2×A + A which becomes eq.6a) D + C = 3×A


  

Hint #6


Substitute A + 2×C for D (from eq.2a) into eq.6a: A + 2×C + C = 3×A which becomes A + 3×C = 3×A Subtract A from each side of the above equation: A + 3×C - A = 3×A - A which makes 3×C = 2×A Divide both sides by 2: 3×C ÷ 2 = 2×A ÷ 2 which makes 1½×C = A and also makes 1½×C = A = B


  

Hint #7


Substitute 1½×C for A in eq.2a: D = 1½×C + 2×C which makes D = 3½×C


  

Solution

Substitute 1½×C for A and B, 3½×C for D, 0 for E, and C for F in eq.1: 1½×C + 1½×C + C + 3½×C + 0 + C = 17 which simplifies to 8½×C = 17 Divide both sides of the above equation by 8½: 8½×C ÷ 8½ = 17 ÷ 8½ which means C = 2 making A = B = 1½×C = 1½ × 2 = 3 D = 3½×C = 3½ × 2 = 7 F = C = 2 and ABCDEF = 332702