Puzzle for January 4, 2021  ( )

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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) B = A + D eq.3) C = B + D eq.4) D + E = A + B eq.5) E = D + F eq.6) C + D = B + 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 B + D (from eq.3): B + D + D = B + F which becomes B + 2×D = B + F Subtract B from both sides of the equation above: B + 2×D – B = B + F – B which makes 2×D = F


  

Hint #2


In eq.5, replace F with 2×D: E = D + 2×D which makes E = 3×D


  

Hint #3


In eq.4, substitute 3×D for E, and A + D for B (from eq.2): D + 3×D = A + A + D which becomes 4×D = 2×A + D Subtract D from each side of the above equation: 4×D – D = 2×A + D – D which makes 3×D = 2×A Divide both sides by 2: 3×D ÷ 2 = 2×A ÷ 2 which makes 1½×D = A


  

Hint #4


Substitute 1½×D for A in eq.2: B = 1½×D + D which makes B = 2½×D


  

Hint #5


Substitute 2½×D for B in eq.3: C = 2½×D + D which makes C = 3½×D


  

Solution

Substitute 1½×D for A, 2½×D for B, 3½×D for C, 3×D for E, and 2×D for F in eq.1: 1½×D + 2½×D + 3½×D + D + 3×D + 2×D = 27 which simplifies to 13½×D = 27 Divide both sides of the equation above by 13½: 13½×D ÷ 13½ = 27 ÷ 13½ which means D = 2 making A = 1½×D = 1½ × 2 = 3 B = 2½×D = 2½ × 2 = 5 C = 3½×D = 3½ × 2 = 7 E = 3×D = 3 × 2 = 6 F = 2×D = 2 × 2 = 4 and ABCDEF = 357264