Puzzle for July 28, 2019  ( )

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

eq.1) A + B + C + D + E + F = 30 eq.2) D = B + C eq.3) E - A = B + C - D eq.4) F - B = C eq.5) B - C + E = C eq.6)* C! = A + D + E + F

A, B, C, D, E, and F each represent a one-digit positive integer.
*  C! is C-factorial.

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Hint #1


In eq.2, replace C with F - B (from eq.4): D = B + F - B which becomes D = F


  

Hint #2


In eq.3, replace B + C with D (from eq.2): E - A = D - D which becomes E - A = 0 Add A to both sides of the above equation: E - A + A = 0 + A which means E = A


  

Hint #3


Add 2×C to both sides of eq.5: B - C + E + 2×C = C + 2×C which becomes B + E + C = 3×C which may be written as E + B + C = 3×C In the above equation, substitute D for B + C (from eq.2): eq.5a) E + D = 3×C   In eq.5a, substitute A for E, and F for D: eq.5b) A + F = 3×C


  

Hint #4


eq.1 may be written as: B + C + E + D + A + F = 30 Substitute 3×C for E + D (from eq.5a) and for A + F (from eq.5b) in the equation above: B + C + 3×C + 3×C = 30 which becomes eq.1a) B + 7×C = 30


  

Hint #5


eq.6 may be written as: C! = E + D + A + F Substitute 3×C for E + D (from eq.5a) and for A + F (from eq.5b) in the above equation: C! = 3×C + 3×C which becomes C! = 6×C which may be written as C × (C - 1)! = 6×C Divide both sides of the above equation by C: C × (C - 1)! ÷ C = 6×C ÷ C which becomes (C - 1)! = 6 which may be written as (C - 1)! = 3 × 2 × 1 which means C - 1 = 3 which makes C = 4


  

Hint #6


Substitute 4 for C in eq.1a: B + 7×4 = 30 which becomes B + 28 = 30 Subtract 28 from both sides: B + 28 - 28 = 30 - 28 which makes B = 2


  

Hint #7


Substitute 2 for B, and 4 for C in eq.2: D = 2 + 4 which makes D = 6 and also makes F = D = 6


  

Solution

Substitute 6 for F, and 4 for C in eq.5b: A + 6 = 3×4 which becomes A + 6 = 12 Subtract 6 from both sides of the above equation: A + 6 - 6 = 12 - 6 which means A = 6 and also means E = A = 6 making ABCDEF = 624666