Puzzle for November 21, 2023  ( )

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

eq.1) A + B + C + D + E + F = 29 eq.2) D = B + E eq.3) B = A + F eq.4) E + F = C - F eq.5) C + E = A - C eq.6) B + C = D + E + 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 D with B + E (from eq.2): B + C = B + E + E + F which becomes B + C = B + 2×E + F Subtract B from each side of the equation above: B + C - B = B + 2×E + F - B which becomes eq.6a) C = 2×E + F


  

Hint #2


In eq.4, replace C with 2×E + F (from eq.6a): E + F = 2×E + F - F which becomes E + F = 2×E Subtract E from both sides of the equation above: E + F - E = 2×E - E which makes F = E


  

Hint #3


In eq.6a, substitute E for F: C = 2×E + E which makes C = 3×E


  

Hint #4


Substitute 3×E for C in eq.5: 3×E + E = A - 3×E which becomes 4×E = A - 3×E Add 3×E to both sides of the equation above: 4×E + 3×E = A - 3×E + 3×E which makes 7×E = A


  

Hint #5


Substitute 7×E for A, and E for F in eq.3: B = 7×E + E which makes B = 8×E


  

Hint #6


Substitute 8×E for B in eq.2: D = 8×E + E which makes D = 9×E


  

Solution

Substitute 7×E for A, 8×E for B, 3×E for C, 9×E for D, and E for F in eq.1: 7×E + 8×E + 3×E + 9×E + E + E = 29 which simplifies to 29×E = 29 Divide both sides of the equation above by 29: 29×E ÷ 29 = 29 ÷ 29 which means E = 1 making A = 7×E = 7 × 1 = 7 B = 8×E = 8 × 1 = 8 C = 3×E = 3 × 1 = 3 D = 9×E = 9 × 1 = 9 F = E = 1 and ABCDEF = 783911