Puzzle for June 16, 2020  ( )

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

eq.1) A + B + C + D + E + F = 32 eq.2) B + E = A + C eq.3) D + E = A eq.4) C + D + F = A + B eq.5) B – F = C eq.6) A – C + D = (A ÷ F) + B + C

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 with B – F (from eq.5): B – F + D + F = A + B which becomes B + D = A + B Subtract B from each side of the above equation: B + D – B = A + B – B which makes D = A


  

Hint #2


In eq.3, replace D with A: A + E = A Subtract A from both sides of the above equation: A + E – A = A – A which makes E = 0


  

Hint #3


In eq.2, substitute 0 for E: B + 0 = A + C which makes eq.2a) B = A + C


  

Hint #4


Substitute A + C for B (from eq.2a) in eq.5: A + C – F = C In the equation above, subtract C from both sides, and add F to both sides: A + C – F – C + F = C – C + F which makes A = F


  

Hint #5


Substitute A for D and F, and A + C for B (from eq.2a) in eq.6: A – C + A = (A ÷ A) + A + C + C which becomes 2×A – C = 1 + A + 2×C In the equation above, add C to both sides, and subtract A from both sides: 2×A – C + C – A = 1 + A + 2×C + C – A which makes A = 1 + 3×C and also makes eq.6a) D = F = A = 1 + 3×C


  

Hint #6


Substitute 1 + 3×C for A (from eq.6a) in eq.2a: B = 1 + 3×C + C which makes eq.2b) B = 1 + 4×C


  

Solution

Substitute 1 + 3×C for A and D and F (from eq.6a), 1 + 4×C for B (from eq.2b), and 0 for E in eq.1: 1 + 3×C + 1 + 4×C + C + 1 + 3×C + 0 + 1 + 3×C = 32 which simplifies to 4 + 14×C = 32 Subtract 4 from both sides of the equation above: 4 + 14×C – 4 = 32 – 4 which makes 14×C = 28 Divide both sides by 14: 14×C ÷ 14 = 28 ÷ 14 which means C = 2 making A = D = F = 1 + 3×C = 1 + 3×2 = 1 + 6 = 7 (from eq.6a) B = 1 + 4×C = 1 + 4×2 = 1 + 8 = 9 (from eq.2b) and ABCDEF = 792707