Puzzle for August 8, 2023  ( )

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


  

Hint #2


In eq.6, replace D with 2×B: 2×B + F = A + B + C Subtract B from each side of the equation above: 2×B + F - B = A + B + C - B which becomes eq.6a) B + F = A + C


  

Hint #3


In eq.3, substitute B + F for A + C (from eq.6a): B + D = B + F Subtract B from both sides of the equation above: B + D - B = B + F - B which makes D = F and also makes 2×B = D = F


  

Hint #4


Substitute 2×B for D and F, and B + C for E (from eq.2) in eq.5: 2×B + 2×B = C + B + C - 2×B which becomes 4×B = 2×C - B Add B to both sides of the above equation: 4×B + B = 2×C - B + B which makes 5×B = 2×C Divide both sides by 2: 5×B ÷ 2 = 2×C ÷ 2 which makes 2½×B = C


  

Hint #5


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


  

Hint #6


Substitute 2×B for D, and 2½×B for C in eq.3: B + 2×B = A + 2½×B which becomes 3×B = A + 2½×B Subtract 2½×B from each side of the equation above: 3×B - 2½×B = A + 2½×B - 2½×B which makes ½×B = A


  

Solution

Substitute ½×B for A, 2½×B for C, 2×B for D and F, and 3½×B for E in eq.1: ½×B + B + 2½×B + 2×B + 3½×B + 2×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 = ½×B = ½ × 2 = 1 C = 2½×B = 2½ × 2 = 5 D = F = 2×B = 2 × 2 = 4 E = 3½×B = 3½ × 2 = 7 and ABCDEF = 125474