Puzzle for May 7, 2019  ( )

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

eq.1) A + B + C + D + E + F = 28 eq.2) D + F = A + E eq.3) A + C = B + D + F eq.4) C = E + F eq.5) B = A + D eq.6) E - F = A - E

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

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Hint #1


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


  

Hint #2


In eq.3a, replace C with E + F (from eq.4): E + F = B + E Subtract E from both sides of the above equation: E + F - E = B + E - E which makes F = B


  

Hint #3


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


  

Hint #4


Substitute 2×D for E, and (A + D) for F (from eq.5a) in eq.6: 2×D - (A + D) = A - 2×D which is equivalent to 2×D - A - D = A - 2×D which becomes D - A = A - 2×D Add both A and 2×D to each side of the above equation: D - A + A + 2×D = A - 2×D + A + 2×D which simplifies to 3×D = 2×A Divide both sides by 2: 3×D ÷ 2 = 2×A ÷ 2 which makes 1½×D = A


  

Hint #5


Substitute 1½×D for A in eq.5: B = 1½×D + D which makes B = 2½×D which also makes F = B = 2½×D


  

Hint #6


Substitute 2×D for E, and 2½×D for F in eq.4: C = 2×D + 2½×D which means C = 4½×D


  

Solution

Substitute 1½×D for A, 2½×D for B and F, 4½×D for C, and 2×D for E in eq.1: 1½×D + 2½×D + 4½×D + D + 2×D + 2½×D = 28 which simplifies to 14×D = 28 Divide both sides by 14: 14×D ÷ 14 = 28 ÷ 14 which means D = 2 making A = 1½×D = 1½ × 2 = 3 B = F = 2½×D = 2½ × 2 = 5 C = 4½×D = 4½ × 2 = 9 E = 2×D = 2 × 2 = 4 and ABCDEF = 359245