Puzzle for October 25, 2022  ( )

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

eq.1) A + B + C + D + E + F = 23 eq.2) A = B + C eq.3) D + F = A + B eq.4) F - E = C - B eq.5) A - D = D - C eq.6) D - B = A + B - C

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


  

Hint #2


In eq.5, replace D with 3×B: A - 3×B = 3×B - C Add 3×B and C to both sides of the equation above: A - 3×B + 3×B + C = 3×B - C + 3×B + C which becomes eq.5a) A + C = 6×B


  

Hint #3


In eq.5a, substitute B + C for A (from eq.2): B + C + C = 6×B which becomes B + 2×C = 6×B Subtract B from each side of the above equation: B + 2×C - B = 6×B - B which makes 2×C = 5×B Divide both sides by 2: 2×C ÷ 2 = 5×B ÷ 2 which makes C = 2½×B


  

Hint #4


Substitute 2½×B for C in eq.2: A = B + 2½×B which makes A = 3½×B


  

Hint #5


Substitute 3×B for D, and 3½×B for A in eq.3: 3×B + F = 3½×B + B which becomes 3×B + F = 4½×B Subtract 3×B from each side of the equation above: 3×B + F - 3×B = 4½×B - 3×B which makes F = 1½×B


  

Hint #6


Substitute 1½×B for F, and 2½×B for C in eq.4: 1½×B - E = 2½×B - B which becomes 1½×B - E = 1½×B Subtract 1½×B from both sides of the equation above: 1½×B - E - 1½×B = 1½×B - 1½×B which makes -E = 0 which means E = 0


  

Solution

Substitute 3½×B for A, 2½×B for C, 3×B for D, 0 for E, and 1½×B for F in eq.1: 3½×B + B + 2½×B + 3×B + 0 + 1½×B = 23 which simplifies to 11½×B = 23 Divide both sides of the above equation by 11½: 11½×B ÷ 11½ = 23 ÷ 11½ which means B = 2 making A = 3½×B = 3½ × 2 = 7 C = 2½×B = 2½ × 2 = 5 D = 3×B = 3 × 2 = 6 F = 1½×B = 1½ × 2 = 3 and ABCDEF = 725603