Puzzle for July 13, 2020  ( )

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

eq.1) A + B + C + D + E + F = 38 eq.2) A + F = B eq.3) C + F = D + E eq.4) F = B – C + E eq.5) B – D = A – F eq.6) (D ÷ A) × E = (B ÷ F) + D

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

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Hint #1


Add C to both sides of eq.4: F + C = B – C + E + C which becomes F + C = B + E which may be written as: C + F = B + E In eq.3, replace C + F with B + E: B + E = D + E Subtract E from both sides of the equation above: B + E – E = D + E – E which makes B = D


  

Hint #2


In eq.5, replace D with B: B – B = A – F which becomes 0 = A – F Add F to both sides of the equation above: 0 + F = A – F + F which makes F = A


  

Hint #3


In eq.2, substitute A for F: A + A = B which makes 2×A = B and also makes D = B = 2×A


  

Hint #4


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


  

Hint #5


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


  

Solution

Substitute 2×A for B and D, 2×A + 1 for C (from eq.3a), 1 + A for E (from eq.6a), and A for F in eq.1: A + 2×A + 2×A + 1 + 2×A + 1 + A + A = 38 which simplifies to 9×A + 2 = 38 Subtract 2 from each side of the above equation: 9×A + 2 – 2 = 38 – 2 which makes 9×A = 36 Divide each side by 9: 9×A ÷ 9 = 36 ÷ 9 which means A = 4 making B = D = 2×A = 2 × 4 = 8 C = 2×A + 1 = 2×4 + 1 = 8 + 1 = 9 (from eq.3a) E = 1 + A = 1 + 4 = 5 (from eq.6a) and ABCDEF = 489854