Puzzle for November 1, 2022  ( )

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

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

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Hint #1


In eq.3, replace C with D + E + F (from eq.2): A - D = D + E + F + D which becomes A - D = 2×D + E + F Add D to both sides of the above equation: A - D + D = 2×D + E + F + D which becomes eq.3a) A = 3×D + E + F


  

Hint #2


In eq.4, replace A with 3×D + E + F (from eq.3a): E + F = 3×D + E + F + D - F which becomes E + F = 4×D + E Subtract E from each side of the equation above: E + F - E = 4×D + E - E which makes F = 4×D


  

Hint #3


In eq.6, substitute 4×D for F: 4×D - D = D + E which becomes 3×D = D + E Subtract D from each side of the equation above: 3×D - D = D + E - D which makes 2×D = E


  

Hint #4


In eq.5, substitute 4×D for F: B - 4×D = D + 4×D which becomes B - 4×D = 5×D Add 4×D to both sides of the above equation: B - 4×D + 4×D = 5×D + 4×D which makes B = 9×D


  

Hint #5


Substitute 2×D for E, and 4×D for F in eq.2: C = D + 2×D + 4×D which makes C = 7×D


  

Hint #6


Substitute 2×D for E, and 4×D for F in eq.3a: A = 3×D + 2×D + 4×D which makes A = 9×D


  

Solution

Substitute 9×D for A and B, 7×D for C, 2×D for E, and 4×D for F in eq.1: 9×D + 9×D + 7×D + D + 2×D + 4×D = 32 which simplifies to 32×D = 32 Divide both sides of the above equation by 32: 32×D ÷ 32 = 32 ÷ 32 which means D = 1 making A = B = 9×D = 9 × 1 = 9 C = 7×D = 7 × 1 = 7 E = 2×D = 2 × 1 = 2 F = 4×D = 4 × 1 = 4 and ABCDEF = 997124