Puzzle for March 30, 2019  ( )

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

eq.1) A + B + C + D + E + F = 34 eq.2) D + E + F = A + B + C eq.3) B + D = A eq.4) E = A + D eq.5) C + D + E = B + F eq.6) A + C = B + E

A, B, C, D, E, and F each represent a one-digit non-negative integer.

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Hint #1


eq.2 may be written as: D + E + F = A + C + B In the above equation, replace A + C with B + E (from eq.6): D + E + F = B + E + B Subtract E from both sides: D + E + F - E = B + E + B - E which becomes eq.2a) D + F = 2×B


  

Hint #2


In eq.2, replace A with B + D (from eq.3): D + E + F = B + D + B + C Subtract D from both sides of the equation above: D + E + F - D = B + D + B + C - D which becomes eq.2b) E + F = 2×B + C


  

Hint #3


In eq.2b, substitute D + F for 2×B (from eq.2a): E + F = D + F + C Subtract F from both sides of the above equation: E + F - F = D + F + C - F which becomes eq.2c) E = D + C


  

Hint #4


In eq.2c, substitute A + D for E (from eq.4): A + D = D + C Subtract D from both sides of the equation above: A + D - D = D + C - D which makes A = C


  

Hint #5


Subtract the left and right sides of eq.5 from the left and right sides of eq.2, respectively: D + E + F - (C + D + E) = A + B + C - (B + F) which is equivalent to D + E + F - C - D - E = A + B + C - B - F which simplifies to F - C = A + C - F Add both C and F to each side of the equation above: F - C + C + F = A + C - F + C + F which becomes eq.2d) 2×F = A + 2×C


  

Hint #6


Substitute A for C in eq.2d: 2×F = A + 2×A which becomes 2×F = 3×A Divide both sides of the above equation by 2: 2×F ÷ 2 = 3×A ÷ 2 which makes eq.2e) F = 1½×A


  

Hint #7


Subtract D from both sides of eq.3: B + D - D = A - D which becomes eq.3a) B = A - D


  

Hint #8


Substitute 1½×A for F, and A for C in eq.2: D + E + 1½×A = A + B + A Subtract 1½×A from both sides of the above equation: D + E + 1½×A - 1½×A = A + B + A - 1½×A which becomes eq.2f) D + E = ½×A + B


  

Hint #9


Substitute A + D for E (from eq.4), and A - D for B (from eq.3a) into eq.2f: D + A + D = ½×A + A - D which becomes 2×D + A = 1½×A - D Add (D - A) to both sides of the above equation: 2×D + A + (D - A) = 1½×A - D + (D - A) which simplifies to 3×D = ½×A Divide both sides by ½: 3×D ÷ ½ = ½×A ÷ ½ which makes 6×D = A which also means C = A = 6×D


  

Hint #10


In eq.2e, substitute 6×D for A: F = 1½×(6×D) which makes F = 9×D


  

Hint #11


In eq.3a, substitute 6×D for A: B = 6×D - D which means B = 5×D


  

Hint #12


In eq.4, substitute 6×D for A: E = 6×D + D which makes E = 7×D


  

Solution

Substitute 6×D for A and C, 5×D for B, 7×D for E, and 9×D for F in eq.1: 6×D + 5×D + 6×D + D + 7×D + 9×D = 34 which becomes 34×D = 34 Divide both sides by 34: 34×D ÷ 34 = 34 ÷ 34 which means D = 1 making A = C = 6×D = 6 × 1 = 6 B = 5×D = 5 × 1 = 5 E = 7×D = 7 × 1 = 7 F = 9×D = 9 × 1 = 9 and ABCDEF = 656179