Puzzle for October 25, 2023  ( )

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

eq.1) A + B + C + D + E + F = 20 eq.2) A = C + D eq.3) B = C + E eq.4) F - C = A - F eq.5) E + F = A + D - F eq.6) B - C + D = C + E + F

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 B with C + E (from eq.3): C + E - C + D = C + E + F which becomes E + D = C + E + F Subtract E from both sides of the above equation: E + D - E = C + E + F - E which becomes eq.6a) D = C + F


  

Hint #2


In eq.2, replace D with C + F (from eq.6a): A = C + C + F which becomes eq.2a) A = 2×C + F


  

Hint #3


In eq.4, substitute 2×C + F for A (from eq.2a): F - C = 2×C + F - F which becomes F - C = 2×C Add C to both sides of the equation above: F - C + C = 2×C + C which makes F = 3×C


  

Hint #4


Substitute 3×C for F in eq.2a: A = 2×C + 3×C which makes A = 5×C


  

Hint #5


Substitute 3×C for F in eq.6a: D = C + 3×C which makes D = 4×C


  

Hint #6


Substitute 3×C for F, 5×C for A, and 4×C for D in eq.5: E + 3×C = 5×C + 4×C - 3×C which becomes E + 3×C = 6×C Subtract 3×C from each side of the equation above: E + 3×C - 3×C = 6×C - 3×C which makes E = 3×C


  

Hint #7


Substitute 3×C for E in eq.3: B = C + 3×C which makes B = 4×C


  

Solution

Substitute 5×C for A, 4×C for B and D, and 3×C for E and F in eq.1: 5×C + 4×C + C + 4×C + 3×C + 3×C = 20 which simplifies to 20×C = 20 Divide both sides of the above equation by 20: 20×C ÷ 20 = 20 ÷ 20 which means C = 1 making A = 5×C = 5 × 1 = 5 B = D = 4×C = 4 × 1 = 4 E = F = 3×C = 3 × 1 = 3 and ABCDEF = 541433