Puzzle for October 7, 2020  ( )

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

eq.1) A + B + C + D + E + F = 15 eq.2) B + F = A + E eq.3) D = B – A eq.4) C – B = A + F eq.5) A – B + C = B + E eq.6) D + E + F = C – A

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

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Hint #1


In eq.6, replace D with B – A (from eq.3): B – A + E + F = C – A Add A to both sides of the above equation: B – A + E + F + A = C – A + A which becomes eq.6a) B + E + F = C


  

Hint #2


In eq.4, replace C with B + E + F: B + E + F – B = A + F which becomes E + F = A + F Subtract F from each side of the above equation: E + F – F = A + F – F which makes E = A


  

Hint #3


In eq.2, substitute A for E: B + F = A + A which becomes eq.2a) B + F = 2×A


  

Hint #4


eq.6a may be re-written as: B + F + E = C In the above equation, substitute 2×A for B + F (from eq.2a), and A for E: 2×A + A = C which makes 3×A = C


  

Hint #5


Substitute 3×A for C, and A for E in eq.5: A – B + 3×A = B + A which becomes 4×A – B = B + A In the equation above, subtract A from both sides, and add B to both sides: 4×A – B – A + B = B + A – A + B which makes 3×A = 2×B Divide both sides by 2: 3×A ÷ 2 = 2×B ÷ 2 which makes 1½×A = B


  

Hint #6


Substitute 1½×A for B in eq.3: D = 1½×A – A which makes D = ½×A


  

Hint #7


Substitute 3×A for C, and 1½×A for B in eq.4: 3×A – 1½×A = A + F which becomes 1½×A = A + F Subtract A from each side of the above equation: 1½×A – A = A + F – A which makes ½×A = F


  

Solution

Substitute 1½×A for B, 3×A for C, ½×A for D and F, and A for E in eq.1: A + 1½×A + 3×A + ½×A + A + ½×A = 15 which simplifies to 7½×A = 15 Divide both sides of the above equation by 7½: 7½×A ÷ 7½ = 15 ÷ 7½ which means A = 2 making B = 1½×A = 1½ × 2 = 3 C = 3×A = 3 × 2 = 6 D = F = ½×A = ½ × 2 = 1 E = A = 2 and ABCDEF = 236121