Puzzle for September 16, 2020  ( )

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

eq.1) A + B + C + D + E + F = 27 eq.2) B + C = D + F eq.3) C + D = E + F eq.4) D + E = A eq.5) A + D = B + C + E eq.6) B – E = A + C + E

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

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Hint #1


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


  

Hint #2


In eq.4, replace A with F + E (from eq.5a): D + E = F + E Subtract E from both sides of the equation above: D + E – E = F + E – E which makes D = F


  

Hint #3


In eq.3, substitute D for F: C + D = E + D Subtract D from each sides of the above equation: C + D – D = E + D – D which makes C = E


  

Hint #4


Substitute C for E in eq.6: B – C = A + C + C which becomes B – C = A + 2×C Add C to both sides of the equation above: B – C + C = A + 2×C + C which becomes eq.6a) B = A + 3×C


  

Hint #5


Substitute A + 3×C for B (from eq.6a), and C for E in eq.5: A + D = A + 3×C + C + C which becomes A + D = A + 5×C Subtract A from each side of the above equation: A + D – A = A + 5×C – A which makes D = 5×C and also makes F = D = 5×C


  

Hint #6


Substitute 5×C for D and F in eq.2: B + C = 5×C + 5×C which becomes B + C = 10×C Subtract C from each side of the equation above: B + C – C = 10×C – C which makes B = 9×C


  

Hint #7


Substitute 9×C for B in eq.6a: 9×C = A + 3×C Subtract 3×C from each side of the above equation: 9×C – 3×C = A + 3×C – 3×C which makes 6×C = A


  

Solution

Substitute 6×C for A, 9×C for B, 5×C for D and F, and C for E in eq.1: 6×C + 9×C + C + 5×C + C + 5×C = 27 which simplifies to 27×C = 27 Divide both sides of the equation above by 27: 27×C ÷ 27 = 27 ÷ 27 which means C = 1 making A = 6×C = 6 × 1 = 6 B = 9×C = 9 × 1 = 9 D = F = 5×C = 5 × 1 = 5 E = C = 1 and ABCDEF = 691515