Puzzle for June 29, 2020  ( )

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

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

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Hint #1


In eq.2, replace F with B + E (from eq.5): E + B + E = B + C which becomes B + 2×E = B + C Subtract B from both sides of the above equation: B + 2×E – B = B + C – B which becomes 2×E = C


  

Hint #2


In eq.6, replace C with 2×E: 2×E + E = D which makes 3×E = D


  

Hint #3


In eq.3, substitute 3×E for D, and B + E for F (from eq.5): B + 3×E = A + B + E Subtract B and E from both sides of the above equation: B + 3×E – B – E = A + B + E – B – E which simplifies to 2×E = A


  

Hint #4


Substitute 2×E for A and C, and 3×E for D in eq.4: 2×E – B + 2×E = B – 3×E which becomes 4×E – B = B – 3×E Add B and 3×E to both sides of the above equation: 4×E – B + B + 3×E = B – 3×E + B + 3×E which makes 7×E = 2×B Divide both sides by 2: 7×E ÷ 2 = 2×B ÷ 2 which makes 3½×E = B


  

Hint #5


Substitute 3½×E for B in eq.5: F = 3½×E + E which makes F = 4½×E


  

Solution

Substitute 2×E for A and C, 3½×E for B, 3×E for D, and 4½×E for F in eq.1: 2×E + 3½×E + 2×E + 3×E + E + 4½×E = 32 which simplifies to 16×E = 32 Divide both sides of the above equation by 16: 16×E ÷ 16 = 32 ÷ 16 which means E = 2 making A = C = 2×E = 2 × 2 = 4 B = 3½×E = 3½ × 2 = 7 D = 3×E = 3 × 2 = 6 F = 4½×E = 4½ + 2 = 9 and ABCDEF = 474629