Puzzle for June 14, 2021  ( )

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

eq.1) A + B + C + D + E + F = 19 eq.2) F = B + C eq.3) D + E = C eq.4) C = A + B eq.5) B + D = A + C eq.6) A + F = B + C + E – A

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 C with A + B (from eq.4): B + D = A + A + B which becomes B + D = 2×A + B Subtract B from each side of the above equation: B + D – B = 2×A + B – B which makes D = 2×A


  

Hint #2


In eq.6, replace B + C with F (from eq.2): A + F = F + E – A In the above equation, subtract F from both sides, and add A to both sides: A + F – F + A = F + E – A – F + A which becomes 2×A = E


  

Hint #3


In eq.3, substitute 2×A for D and E: 2×A + 2×A = C which makes 4×A = C


  

Hint #4


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


  

Hint #5


Substitute 3×A for B, and 4×A for C in eq.2: F = 3×A + 4×A which becomes F = 7×A


  

Solution

Substitute 3×A for B, 4×A for C, 2×A for D and E, and 7×A for F in eq.1: A + 3×A + 4×A + 2×A + 2×A + 7×A = 19 which simplifies to 19×A = 19 Divide both sides of the above equation by 19: 19×A ÷ 19 = 19 ÷ 19 which means A = 1 making B = 3×A = 3 × 1 = 3 C = 4×A = 4 × 1 = 4 D = E = 2×A = 2 × 1 = 2 F = 7×A = 7 × 1 = 7 and ABCDEF = 134227