Puzzle for June 5, 2023  ( )

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

eq.1) A + B + C + D + E + F = 31 eq.2) B = C + E eq.3) C = D + F eq.4) E - F = D - E eq.5) D - F = C - D eq.6) C + D = A + E

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 C with D + F (from eq.3): D - F = D + F - D which becomes D - F = F Add F to both sides of the above equation: D - F + F = F + F which makes D = 2×F


  

Hint #2


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


  

Hint #3


In eq.4, replace D with 2×F: E - F = 2×F - E Add F and E to both sides of the above equation: E - F + F + E = 2×F - E + F + E which becomes 2×E = 3×F Divide both sides by 2: 2×E ÷ 2 = 3×F ÷ 2 which makes E = 1½×F


  

Hint #4


In eq.2, substitute 3×F for C, and 1½×F for E: B = 3×F + 1½×F which makes B = 4½×F


  

Hint #5


Substitute 3×F for C, 2×F for D, and 1½×F for E in eq.6: 3×F + 2×F = A + 1½×F which becomes 5×F = A + 1½×F Subtract 1½×F from each side of the equation above: 5×F - 1½×F = A + 1½×F - 1½×F which makes 3½×F = A


  

Solution

Substitute 3½×F for A, 4½×F for B, 3×F for C, 2×F for D, and 1½×F for E in eq.1: 3½×F + 4½×F + 3×F + 2×F + 1½×F + F = 31 which simplifies to 15½×F = 31 Divide both sides of the above equation by 15½: 15½×F ÷ 15½ = 31 ÷ 15½ which means F = 2 making A = 3½×F = 3½ × 2 = 7 B = 4½×F = 4½ × 2 = 9 C = 3×F = 3 × 2 = 6 D = 2×F = 2 × 2 = 4 E = 1½×F = 1½ × 2 = 3 and ABCDEF = 796432