Puzzle for November 30, 2022  ( )

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

eq.1) A + B + C + D + E + F = 28 eq.2) A + B = E + F eq.3) B - F = C + F eq.4) E - F = B + F eq.5) C - D = A - F eq.6) B - C + E = A + C + D

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

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Hint #1


Add F to both sides of eq.4: E - F + F = B + F + F which becomes eq.4a) E = B + 2×F   In eq.2, replace E with B + 2×F (from eq.4a): A + B = B + 2×F + F which becomes A + B = B + 3×F Subtract B from each side of the equation above: A + B - B = B + 3×F - B which makes A = 3×F


  

Hint #2


Add F to both sides of eq.3: B - F + F = C + F + F which becomes eq.3a) B = C + 2×F   In eq.4a, replace B with C + 2×F (from eq.3a): E = C + 2×F + 2×F which becomes eq.4b) E = C + 4×F


  

Hint #3


In eq.6, replace B with C + 2×F (from eq.3a), E with C + 4×F (from eq.4b), and A with 3×F: C + 2×F - C + C + 4×F = 3×F + C + D which becomes C + 6×F = 3×F + C + D Subtract C and 3×F from each side of the above equation: C + 6×F - C - 3×F = 3×F + C + D - C - 3×F which simplifies to 3×F = D


  

Hint #4


In eq.5, substitute 3×F for D and A: C - 3×F = 3×F - F which becomes C - 3×F = 2×F Add 3×F to both sides of the above equation: C - 3×F + 3×F = 2×F + 3×F which makes C = 5×F


  

Hint #5


Substitute 5×F for C in eq.3a: B = 5×F + 2×F which makes B = 7×F


  

Hint #6


Substitute 5×F for C in eq.4b: E = 5×F + 4×F which makes E = 9×F


  

Solution

Substitute 3×F for A and D, 7×F for B, 5×F for C, and 9×F for E in eq.1: 3×F + 7×F + 5×F + 3×F + 9×F + F = 28 which simplifies to 28×F = 28 Divide both sides of the above equation by 28: 28×F ÷ 28 = 28 ÷ 28 which means F = 1 making A = D = 3×F = 3 × 1 = 3 B = 7×F = 7 × 1 = 7 C = 5×F = 5 × 1 = 5 E = 9×F = 9 × 1 = 9 and ABCDEF = 375391