Puzzle for June 21, 2019  ( )

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

eq.1) A + B + C + D + E + F = 25 eq.2) B + C = D + E eq.3) A + C = B + D eq.4) D + F = A + E eq.5) A - B + F = B + D + E eq.6)* B + C = DE - EF

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  DE and EF are 2-digit numbers (not D×E or E×F).

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Hint #1


In eq.5, replace B + D with A + C (from eq.3): A - B + F = A + C + E Add (B - A) to both sides of the above equation: A - B + F + (B - A) = A + C + E + (B - A) which becomes eq.5a) F = C + E + B


  

Hint #2


In eq.4, replace F with C + E + B (from eq.5a): D + C + E + B = A + E Subtract E from both sides of the above equation: D + C + E + B - E = A + E - E which becomes eq.2a) D + C + B = A


  

Hint #3


In eq.3, replace A with D + C + B (from eq.2a): D + C + B + C = B + D Subtract both B and D from each side of the equation above: D + C + B + C - B - D = B + D - B - D which simplifies to 2×C = 0 which means C = 0


  

Hint #4


Substitute 0 for C in eq.3: A + 0 = B + D which means eq.3a) A = B + D   Substitute 0 for C in eq.2: B + 0 = D + E which means eq.2b) B = D + E   Substitute 0 for C in eq.5a: F = 0 + E + B which makes eq.5b) F = E + B


  

Hint #5


eq.6 may be written as: B + C = 10×D + E - (10×E + F) which is equivalent to B + C = 10×D + E - 10×E - F Substitute 0 for C, and (E + B) for F (from eq.5a) in the above equation: B + 0 = 10×D + E - 10×E - (E + B) which becomes B = 10×D - 9×E - E - B which becomes B = 10×D - 10×E - B Add B to both sides: B + B = 10×D - 10×E - B + B which becomes 2×B = 10×D - 10×E Divide both sides by 2: 2×B ÷ 2 = (10×D - 10×E) ÷ 2 which means eq.6a) B = 5×D - 5×E


  

Hint #6


Substitute (D + E) for B (from eq.2b) in eq.6a: D + E = 5×D - 5×E Add (5×E - D) to both sides of the above equation: D + E + (5×E - D) = 5×D - 5×E + (5×E - D) which simplifies to 6×E = 4×D Divide both sides by 4: 6×E ÷ 4 = 4×D ÷ 4 which makes 1½×E = D


  

Hint #7


Substitute 1½×E for D in eq.2b: B = 1½×E + E which makes B = 2½×E


  

Hint #8


Substitute 2½×E for B in eq.5b: F = E + 2½×E which makes F = 3½×E


  

Hint #9


Substitute 2½×E for B, and 1½×E for D in eq.3a: A = 2½×E + 1½×E which makes A = 4×E


  

Solution

Substitute 4×E for A, 2½×E for B, 0 for C, 1½×E for D, and 3½×E for F in eq.1: 4×E + 2½×E + 0 + 1½×E + E + 3½×E = 25 which simplifies to 12½×E = 25 Divide both sides of the equation above by 12½: 12½×E ÷ 12½ = 25 ÷ 12½ which means E = 2 making A = 4×E = 4 × 2 = 8 B = 2½×E = 2½ × 2 = 5 D = 1½×E = 1½ × 2 = 3 F = 3½×E = 3½ × 2 = 7 and ABCDEF = 850327