Puzzle for May 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 = 28 eq.2) A + B = F eq.3) B + C = E eq.4) C + D = A eq.5) B + D = C + E eq.6)* AB - DE = F

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

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Hint #1


In eq.5, replace E with B + C (from eq.3): B + D = C + B + C Subtract B from both sides of the above equation: B + D - B = C + B + C - B which makes D = 2×C


  

Hint #2


In eq.4, replace D with 2×C: C + 2×C = A which means 3×C = A


  

Hint #3


eq.6 may be written as: 10×A + B - (10×D + E) = F which is equivalent to 10×A + B - 10×D - E = F In the above equation, replace F with A + B (from eq.2): 10×A + B - 10×D - E = A + B Subtract both A and B from each side: 10×A + B - 10×D - E - A - B = A + B - A - B which becomes eq.6a) 9×A - 10×D - E = 0


  

Hint #4


In eq.6a, replace A with (3×C), and D with (2×C): 9×(3×C) - 10×(2×C) - E = 0 which becomes 27×C - 20×C - E = 0 which means 7×C - E = 0 Add E to both sides of the above equation: 7×C - E + E = 0 + E which makes 7×C = E


  

Hint #5


In eq.3, substitute 7×C for E: B + C = 7×C Subtract C from each side: B + C - C = 7×C - C which makes B = 6×C


  

Hint #6


Substitute 3×C for A, and 6×C for B in eq.2: 3×C + 6×C = F which makes 9×C = F


  

Solution

Substitute 3×C for A, 6×C for B, 2×C for D, 7×C for E, and 9×C for F in eq.1: 3×C + 6×C + C + 2×C + 7×C + 9×C = 28 which simplifies to 28×C = 28 Divide both sides by 28: 28×C ÷ 28 = 28 ÷ 28 which means C = 1 making A = 3×C = 3 × 1 = 3 B = 6×C = 6 × 1 = 6 D = 2×C = 2 × 1 = 2 E = 7×C = 7 × 1 = 7 F = 9×C = 9 × 1 = 9 and ABCDEF = 361279