Puzzle for June 9, 2022  ( )

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

eq.1) A + B + C + D + E + F = 27 eq.2) C = D + F eq.3) F = D + E eq.4) B + E = C + D eq.5) E + F = B + D eq.6) C + D = A + F

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

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Hint #1


In eq.6, replace C with D + F (from eq.2): D + F + D = A + F which becomes 2×D + F = A + F Subtract F from each side of the above equation: 2×D + F – F = A + F – F which makes 2×D = A


  

Hint #2


In eq.5, replace F with D + E (from eq.3): E + D + E = B + D which becomes 2×E + D = B + D Subtract D from each side of the equation above: 2×E + D – D = B + D – D which makes 2×E = B


  

Hint #3


In eq.4, substitute 2×E for B: 2×E + E = C + D which becomes eq.4a) 3×E = C + D


  

Hint #4


Substitute 3×E for C + D (from eq.4a) in eq.6: eq.6a) 3×E = A + F


  

Hint #5


eq.1 may be written as: A + F + C + D + B + E = 27 Substitute 3×E for A + F (from eq.6a), 3×E for C + D (from eq.4a), and 2×E for B in the above equation: 3×E + 3×E + 2×E + E = 27 which becomes 9×E = 27 Divide both sides by 9: 9×E ÷ 9 = 27 ÷ 9 which makes E = 3 and makes B = 2×E = 2 × 3 = 6


  

Hint #6


Substitute 3 for E in eq.3: eq.3a) F = D + 3


  

Hint #7


Substitute 3 for E, 2×D for A, and D + 3 for F (from eq.3a) in eq.6a: 3×3 = 2×D + D + 3 which becomes 9 = 3×D + 3 Subtract 3 from both sides of the above equation: 9 – 3 = 3×D + 3 – 3 which makes 6 = 3×D Divide both sides by 3: 6 ÷ 3 = 3×D ÷ 3 which makes 2 = D and makes A = 2×D = 2 × 2 = 4


  

Hint #8


Substitute 2 for D, and 3 for E in eq.3: F = 2 + 3 which makes F = 5


  

Solution

Substitute 2 for D, and 5 for F in eq.2: C = 2 + 5 which makes C = 7 and makes ABCDEF = 467235