Puzzle for March 6, 2023  ( )

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

eq.1) A + B + C + D + E + F = 27 eq.2) E = B + C eq.3) A = C + E eq.4) D = B + E eq.5) E + F = A + C eq.6) B - C = C + F

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 A with C + E (from eq.3): E + F = C + E + C which becomes E + F = 2×C + E Subtract E from each side of the equation above: E + F - E = 2×C + E - E which makes F = 2×C


  

Hint #2


In eq.6, replace F with 2×C: B - C = C + 2×C which becomes B - C = 3×C Add C to both sides of the above equation: B - C + C = 3×C + C which makes B = 4×C


  

Hint #3


In eq.2, substitute 4×C for B: E = 4×C + C which makes E = 5×C


  

Hint #4


Substitute 5×C for E in eq.3: A = C + 5×C which makes A = 6×C


  

Hint #5


Substitute 4×C for B, and 5×C for E in eq.4: D = 4×C + 5×C which makes D = 9×C


  

Solution

Substitute 6×C for A, 4×C for B, 9×C for D, 5×C for E, and 2×C for F in eq.1: 6×C + 4×C + C + 9×C + 5×C + 2×C = 27 which simplifies to 27×C = 27 Divide both sides of the above equation by 27: 27×C ÷ 27 = 27 ÷ 27 which means C = 1 making A = 6×C = 6 × 1 = 6 B = 4×C = 4 × 1 = 4 D = 9×C = 9 × 1 = 9 E = 5×C = 5 × 1 = 5 F = 2×C = 2 × 1 = 2 and ABCDEF = 641952