Puzzle for September 29, 2020  ( )

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

eq.1) A + B + C + D + E + F = 33 eq.2) D + F = A + B + C eq.3) C + E = A + B – C eq.4) F – C = B eq.5) E – D = A + D – E eq.6) A ÷ D = C

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

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Hint #1


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


  

Hint #2


In eq.6, replace D with A: A ÷ A = C which means 1 = C


  

Hint #3


Add D and E to both sides of eq.5: E – D + D + E = A + D – E + D + E which becomes 2×E = A + 2×D In the above equation, replace D with A: 2×E = A + 2×A which means 2×E = 3×A Divide both sides by 2: 2×E ÷ 2 = 3×A ÷ 2 which makes E = 1½×A


  

Hint #4


Substitute 1 for C, and 1½×A for E in eq.3: 1 + 1½×A = A + B – 1 In the equation above, add 1 to both sides, and subtract A from both sides: 1 + 1½×A + 1 – A = A + B – 1 + 1 – A which makes eq.3a) 2 + ½×A = B


  

Hint #5


Substitute 1 for C, and 2 + ½×A for B (from eq.3a) in eq.4: F – 1 = 2 + ½×A Add 1 to both sides: F – 1 + 1 = 2 + ½×A + 1 which becomes eq.4a) F = 3 + ½×A


  

Solution

Substitute 2 + ½×A for B (from eq.3a), 1 for C, A for D, 1½×A for E, and 3 + ½×A for F (from eq.4a) in eq.1: A + 2 + ½×A + 1 + A + 1½×A + 3 + ½×A = 33 which simplifies to 4½×A + 6 = 33 Subtract 6 from both sides of the equation above: 4½×A + 6 – 6 = 33 – 6 which makes 4½×A = 27 Divide both sides by 4½: 4½×A ÷ 4½ = 27 ÷ 4½ which means A = 6 making B = 2 + ½×A = 2 + ½×6 = 2 + 3 = 5 (from eq.3a) D = A = 6 E = 1½×A = 1½ × 6 = 9 F = 3 + ½×A = 3 + ½×6 = 3 + 3 = 6 (from eq.4a) and ABCDEF = 651696