Puzzle for January 8, 2020  ( )

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

eq.1) A + B + C + D + E + F = 21 eq.2) C + F = A eq.3) E = A + C eq.4) B – D = C – F eq.5) A + D = E – F eq.6) E – D = A + B

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 E with A + C (from eq.3): A + D = A + C – F Subtract A from each side of the equation above: A + D – A = A + C – F – A which becomes eq.5a) D = C – F


  

Hint #2


In eq.4, replace C – F with D (from eq.5a): B – D = D Add D to each side of the above equation: B – D + D = D + D which makes B = 2×D


  

Hint #3


In eq.6, substitute 2×D for B: E – D = A + 2×D Add D to both sides of the equation above: E – D + D = A + 2×D + D which becomes eq.6a) E = A + 3×D


  

Hint #4


Substitute A + 3×D for E (from eq.6a) in eq.3: A + 3×D = A + C Subtract A from both sides of the above equation: A + 3×D – A = A + C – A which makes 3×D = C


  

Hint #5


Substitute 2×D for B, and 3×D for C in eq.4: 2×D – D = 3×D – F which becomes D = 3×D – F In the equation above, add F to both sides, and subtract D from each side: D + F – D = 3×D – F + F – D which makes F = 2×D


  

Hint #6


Substitute 3×D for C, and 2×D for F in eq.2: 3×D + 2×D = A which makes 5×D = A


  

Hint #7


Substitute 5×D for A in eq.6a: E = 5×D + 3×D which makes E = 8×D


  

Solution

Substitute 5×D for A, 2×D for B and F, 3×D for C, and 8×D for E in eq.1: 5×D + 2×D + 3×D + D + 8×D + 2×D = 21 which simplifies to 21×D = 21 Divide both sides of the above equation by 21: 21×D ÷ 21 = 21 ÷ 21 which means D = 1 making A = 5×D = 5 × 1 = 5 B = F = 2×D = 2 × 1 = 2 C = 3×D = 3 × 1 = 3 E = 8×D = 8 × 1 = 8 and ABCDEF = 523182