Puzzle for May 4, 2023  ( )

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

eq.1) A + B + C + D + E + F = 30 eq.2) F = A + D eq.3) F = B + C eq.4) B + E = C + F eq.5) C + D + F = A + B + E eq.6) B = average (C, D, F)

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 B + E with C + F (from eq.4): C + D + F = A + C + F Subtract C and F from each side of the above equation: C + D + F - C - F = A + C + F - C - F which simplifies to D = A


  

Hint #2


In eq.2, replace D with A: F = A + A which makes F = 2×A


  

Hint #3


In eq.4, substitute B + C for F (from eq.3): B + E = C + B + C which becomes B + E = 2×C + B Subtract B from both sides of the above equation: B + E - B = 2×C + B - B which makes E = 2×C


  

Hint #4


eq.6 may be written as: B = (C + D + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × B = 3 × (C + D + F) ÷ 3 which becomes eq.6a) 3×B = C + D + F


  

Hint #5


eq.1 may be written as: A + B + E + C + D + F = 30 In the equation above, replace A + B + E with C + D + F (from eq.5): C + D + F + C + D + F = 30 which becomes 2×(C + D + F) = 30 Divide both sides by 2: 2×(C + D + F) ÷ 2 = 30 ÷ 2 which becomes eq.1a) C + D + F = 15


  

Hint #6


In eq.1a, substitute 3×B for C + D + F (from eq.6a): 3×B = 15 Divide both sides of the above equation by 3: 3×B ÷ 3 = 15 ÷ 3 which makes B = 5


  

Hint #7


Substitute 5 for B, A for D, and 2×A for F in eq.6a: 3×5 = C + A + 2×A which becomes eq.6b) 15 = C + 3×A


  

Hint #8


Substitute 15 for C + D + F (from eq.1a), 5 for B, and 2×C for E in eq.5: 15 = A + 5 + 2×C Subtract 5 and 2×C from each side of the equation above: 15 - 5 - 2×C = A + 5 + 2×C - 5 - 2×C which becomes eq.5a) 10 - 2×C = A


  

Hint #9


Substitute (10 - 2×C) for A (from eq.5a) into eq.6b: 15 = C + 3×(10 - 2×C) which becomes 15 = C + 30 - 6×C which makes 15 = 30 - 5×C In the equation above, subtract 15 from both sides, and add 5×C to both sides: 15 - 15 + 5×C = 30 - 5×C - 15 + 5×C which makes 5×C = 15 Divide both sides by 5: 5×C ÷ 5 = 15 ÷ 5 which makes C = 3 making E = 2×C = 2×3 = 6


  

Solution

Substitute 3 for C in eq.5a: 10 - 2×3 = A which becomes 10 - 6 = A which makes 4 = A making D = A = 4 F = 2×A = 2×4 = 8 and ABCDEF = 453468