Puzzle for January 5, 2020  ( )

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

eq.1) A + B + C + D + E + F = 35 eq.2) A = C + D eq.3) B + F = A + C eq.4) C + D + F = A + B eq.5) D + E = A + B + C eq.6) E + F = A + D

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

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Hint #1


In eq.4, replace C + D with A (from eq.2): A + F = A + B Subtract A from both sides of the above equation: A + F – A = A + B – A which makes F = B


  

Hint #2


In eq.3, replace F with B: B + B = A + C which becomes eq.3a) 2×B = A + C


  

Hint #3


eq.5 may be written as: D + E = A + C + B In the equation above, substitute 2×B for A + C (from eq.3a): D + E = 2×B + B which becomes eq.5a) D + E = 3×B


  

Hint #4


eq.1 may be written as: A + C + B + D + E + F = 35 Substitute 2×B for A + C (from eq.3a), 3×B for D + E (from eq.5a), and B for F in eq.1: 2×B + B + 3×B + B = 35 which becomes 7×B = 35 Divide both sides of the equation above by 7: 7×B ÷ 7 = 35 ÷ 7 which makes B = 5 and also makes F = B = 5


  

Hint #5


Substitute 5 for B in eq.5a: D + E = 3×5 which becomes D + E = 15 Subtract E from each side of the equation above: D + E – E = 15 – E which becomes eq.5b) D = 15 – E


  

Hint #6


Substitute 5 for both B and F, and C + D for A (from eq.2) in eq.3: 5 + 5 = C + D + C which becomes eq.3b) 10 = 2×C + D


  

Hint #7


Substitute 15 – E for D (from eq.5b) in eq.3b: 10 = 2×C + 15 – E In the above equation, add E to both sides, and subtract 15 from each side: 10 + E – 15 = 2×C + 15 – E + E – 15 which becomes eq.3c) E – 5 = 2×C


  

Hint #8


Substitute 5 for F, and 15 – E for D (from eq.5b) in eq.6: E + 5 = A + 15 – E In the above equation, add E to each side, and subtract 15 from both sides: E + 5 + E – 15 = A + 15 – E + E – 15 which becomes 2×E – 10 = A Divide both sides by 2: (2×E – 10) ÷ 2 = A ÷ 2 which becomes eq.6a) E – 5 = ½×A


  

Hint #9


Substitute ½×A for E – 5 (from eq.6a) in eq.3c: ½×A = 2×C Multiply each side of the above equation by 2: ½×A × 2 = 2×C × 2 which makes A = 4×C


  

Hint #10


Substitute 5 for B, and 4×C for A in eq.3a: 2×5 = 4×C + C which makes 10 = 5×C Divide both sides of the equation above by 5: 10 ÷ 5 = 5×C ÷ 5 which makes 2 = C making A = 4×C = 4 × 2 = 8


  

Solution

Substitute 2 for C in eq.3c: E – 5 = 2×2 which becomes E – 5 = 4 Add 5 to both sides of the equation above: E – 5 + 5 = 4 + 5 which makes E = 9 making D = 15 – E = 15 – 9 = 6 (from eq.5b) and ABCDEF = 852695