Puzzle for October 27, 2021  ( )

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

eq.1) A + B + C + D + E + F = 32 eq.2) F = B + D eq.3) D + F = A eq.4) C + D = B eq.5) B + D + E = C + F eq.6) A – C + F = C + E

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

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Hint #1


In eq.2, replace B with C + D (from eq.4): F = C + D + D which becomes eq.2a) F = C + 2×D


  

Hint #2


In eq.5, replace B with C + D (from eq.4): C + D + D + E = C + F which becomes C + 2×D + E = C + F Subtract C from each side of the above equation: C + 2×D + E – C = C + F – C which becomes eq.5a) 2×D + E = F


  

Hint #3


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


  

Hint #4


Substitute C + 2×D for F (from eq.2a) in eq.3: D + C + 2×D = A which becomes eq.3a) C + 3×D = A


  

Hint #5


Substitute C + 3×D for A (from eq.3a), C + 2×D for F (from eq.2a), and C for E in eq.6: C + 3×D – C + C + 2×D = C + C which becomes C + 5×D = 2×C Subtract C from both sides of the above equation: C + 5×D – C = 2×C – C which makes 5×D = C and also makes E = C = 5×D


  

Hint #6


Substitute 5×D for C in eq.3a: 5×D + 3×D = A which makes 8×D = A


  

Hint #7


Substitute 5×D for E in eq.5a: 2×D + 5×D = F which makes 7×D = F


  

Hint #8


Substitute 7×D for F in eq.2: 7×D = B + D Subtract D from each side of the equation above: 7×D – D = B + D – D which makes 6×D = B


  

Solution

Substitute 8×D for A, 6×D for B, 5×D for C and E, and 7×D for F in eq.1: 8×D + 6×D + 5×D + D + 5×D + 7×D = 32 which simplifies to 32×D = 32 Divide both sides of the equation above by 32: 32×D ÷ 32 = 32 ÷ 32 which means D = 1 making A = 8×D = 8 × 1 = 8 B = 6×D = 6 × 1 = 6 C = E = 5×D = 5 × 1 = 5 F = 7×D = 7 × 1 = 7 and ABCDEF = 865157