Puzzle for December 12, 2023  ( )

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

eq.1) A + B + C + D + E + F = 35 eq.2) F = A + B eq.3) C = B + E eq.4) D + E = A + C eq.5) C + E = F - E eq.6) B + C = 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 D + E with A + C (from eq.4): B + C = A + A + C which becomes B + C = 2×A + C Subtract C from each side of the equation above: B + C - C = 2×A + C - C which makes eq.6a) B = 2×A


  

Hint #2


In eq.2, replace B with 2×A: F = A + 2×A which makes eq.2a) F = 3×A


  

Hint #3


In eq.3, replace B with 2×A: eq.3a) C = 2×A + E


  

Hint #4


In eq.5, substitute 2×A + E for C (from eq.3a), and 3×A for F: 2×A + E + E = 3×A - E which becomes 2×A + 2×E = 3×A - E In the above equation, subtract 2×A from both sides, and add E to both sides: 2×A + 2×E - 2×A + E = 3×A - E - 2×A + E which makes 3×E = A


  

Hint #5


Substitute (3×E) for A in eq.3a: C = 2×(3×E) + E which becomes C = 6×E + E which makes C = 7×E


  

Hint #6


Substitute (3×E) for A in eq.2a: F = 3×(3×E) which makes F = 9×E


  

Hint #7


Substitute (3×E) for A in eq.6a: B = 2×(3×E) which makes B = 6×E


  

Hint #8


Substitute 3×E for A, and 7×E for C in eq.4: D + E = 3×E + 7×E which becomes D + E = 10×E Subtract E from each side of the above equation: D + E - E = 10×E - E which makes D = 9×E


  

Solution

Substitute 3×E for A, 6×E for B, 7×E for C, and 9×E for D and F in eq.1: 3×E + 6×E + 7×E + 9×E + E + 9×E = 35 which simplifies to 35×E = 35 Divide both sides of the above equation by 35: 35×E ÷ 35 = 35 ÷ 35 which means E = 1 making A = 3×E = 3 × 1 = 3 B = 6×E = 6 × 1 = 6 C = 7×E = 7 × 1 = 7 D = F = 9×E = 9 × 1 = 9 and ABCDEF = 367919