Puzzle for May 31, 2023  ( )

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

eq.1) F = B + E eq.2) C - D = D - B eq.3) B + C = A - B + F eq.4) E - D = D + F - C eq.5) A + D = E + F - A eq.6) E - A = C ÷ 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 F with B + E (from eq.1): E - D = D + B + E - C In the above equation, subtract E from both sides, and add D and C to both sides: E - D - E + D + C = D + B + E - C - E + D + C which simplifies to eq.4a) C = 2×D + B


  

Hint #2


Add D and B to both sides of eq.2: C - D + D + B = D - B + D + B which becomes eq.2a) C + B = 2×D


  

Hint #3


In eq.4a, replace 2×D with C + B (from eq.2a): C = C + B + B which becomes C = C + 2×B Subtract C from each side of the equation above: C - C = C + 2×B - C which makes 0 = 2×B which means 0 = B


  

Hint #4


In eq.4a, substitute 0 for B: C = 2×D + 0 which makes C = 2×D


  

Hint #5


In eq.1, substitute 0 for B: F = 0 + E which makes F = E


  

Hint #6


Substitute 2×D for C in eq.6: E - A = 2×D ÷ D which becomes E - A = 2 Add A to both sides of the above equation: E - A + A = 2 + A which makes E = 2 + A and also makes eq.6a) F = E = 2 + A


  

Hint #7


Substitute 0 for B, 2×D for C, and 2 + A for F (from eq.6a) in eq.3: 0 + 2×D = A - 0 + 2 + A which makes 2×D = 2×A + 2 Divide both sides of the above equation by 2: 2×D ÷ 2 = (2×A + 2) ÷ 2 which makes eq.3a) D = A + 1


  

Hint #8


Substitute A + 1 for D (from eq.3a), and 2 + A for E and F (from eq.6a) in eq.5: A + A + 1 = 2 + A + 2 + A - A which becomes 2×A + 1 = 4 + A Subtract 1 and A from each side of the equation above: 2×A + 1 - 1 - A = 4 + A - 1 - A which makes A = 3


  

Solution

Since A = 3, then: E = F = 2 + 3 = 5 (from eq.6a) D = A + 1 = 3 + 1 = 4 (from eq.3a) C = 2×D = 2 × 4 = 8 and ABCDEF = 308455