Puzzle for August 18, 2021  ( )

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

eq.1) A + B + C + D + E + F = 34 eq.2) B = A + C eq.3) D + E = F eq.4) A – C + D = B + C eq.5) B + E + F = A + C + D eq.6) A = average (B, C, D, E)

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 B with A + C (from eq.2): A – C + D = A + C + C which becomes A – C + D = A + 2×C In the above equation, subtract A from both sides, and add C to both sides: A – C + D – A + C = A + 2×C – A + C which makes D = 3×C


  

Hint #2


In eq.5, replace A + C with B (from eq.2): B + E + F = B + D Subtract B from each side of the above equation: B + E + F – B = B + D – B which becomes eq.5a) E + F = D


  

Hint #3


In eq.3, substitute E + F for D (from eq.5a): E + F + E = F which becomes 2×E + F = F Subtract F from each side of the above equation: 2×E + F – F = F – F which makes 2×E = 0 which means E = 0


  

Hint #4


Substitute 0 for E in eq.5a: 0 + F = D which makes F = D and also makes F = D = 3×C


  

Hint #5


eq.6 may be written as: A = (B + C + D + E) ÷ 4 Multiply both sides of the equation above by 4: 4 × A = 4 × (B + C + D + E) ÷ 4 which becomes 4×A = B + C + D + E which may be written as eq.6a) 4×A = B + D + C + E


  

Hint #6


Substitute A + C for B (from eq.2), 3×C for D, and 0 for E in eq.6a: 4×A = A + C + 3×C + C + 0 which becomes 4×A = A + 5×C Subtract A from each side of the above equation: 4×A – A = A + 5×C – A which becomes 3×A = 5×C Divide both sides by 3: 3×A ÷ 3 = 5×C ÷ 3 which makes A = 1⅔×C


  

Hint #7


Substitute 1⅔×C for A in eq.2: B = 1⅔×C + C which makes B = 2⅔×C


  

Solution

Substitute 1⅔×C for A, 2⅔×C for B, 3×C for D and F, and 0 for E in eq.1: 1⅔×C + 2⅔×C + C + 3×C + 0 + 3×C = 34 which simplifies to 11⅓×C = 34 Divide both sides of the above equation by 11⅓: 11⅓×C ÷ 11⅓ = 34 ÷ 11⅓ which means C = 3 making A = 1⅔×C = 1⅔ × 3 = 5 B = 2⅔×C = 2⅔ × 3 = 8 D = F = 3×C = 3 × 3 = 9 and ABCDEF = 583909