Puzzle for September 10, 2023  ( )

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

eq.1) A + B + C + D + E + F = 24 eq.2) D = B + F eq.3) B - D = D - A eq.4) C + E = D + F eq.5) E - D = C + D - B eq.6) B = average (A, C, F)

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

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Hint #1


Subtract B from each side of eq.2: D - B = B + F - B which becomes eq.2a) D - B = F


  

Hint #2


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


  

Hint #3


Add D and B to both sides of eq.5: E - D + D + B = C + D - B + D + B which becomes eq.5a) E + B = C + 2×D


  

Hint #4


In eq.4a, replace E + B with C + 2×D (from eq.5a): C + C + 2×D = 2×D which becomes 2×C + 2×D = 2×D Subtract 2×D from each side of the equation above: 2×C + 2×D - 2×D = 2×D - 2×D which makes 2×C = 0 which means C = 0


  

Hint #5


In eq.5a, substitute 0 for C: E + B = 0 + 2×D which becomes eq.5b) E + B = 2×D


  

Hint #6


Add D and A to both sides of eq.3: B - D + D + A = D - A + D + A which becomes eq.3a) B + A = 2×D


  

Hint #7


Substitute E + B for 2×D (from eq.5b) into eq.3a: B + A = E + B Subtract B from both sides of the equation above: B + A - B = E + B - B which makes A = E


  

Hint #8


eq.6 may be written as: B = (A + C + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × B = 3 × (A + C + F) ÷ 3 which becomes eq.6a) 3×B = A + C + F


  

Hint #9


Substitute E for A, and 0 for C in eq.6a: 3×B = E + 0 + F which becomes eq.6b) 3×B = E + F


  

Hint #10


Subtract F from each side of eq.2: D - F = B + F - F which becomes eq.2b) D - F = B


  

Hint #11


Substitute 0 for C in eq.4: 0 + E = D + F which becomes eq.4b) E = D + F


  

Hint #12


Substitute D - F for B (from eq.2b), and D + F for E (from eq.4b) in eq.6b: 3×(D - F) = D + F + F which becomes 3×D - 3×F = D + 2×F In the equation above, add 3×F to both sides, and subtract D from both sides: 3×D - 3×F + 3×F - D = D + 2×F + 3×F - D which becomes 2×D = 5×F Divide both sides by 2: 2×D ÷ 2 = 5×F ÷ 2 which makes D = 2½×F


  

Hint #13


Substitute 2½×F for D in eq.4b: E = 2½×F + F which makes E = 3½×F and also makes A = E = 3½×F


  

Hint #14


Substitute 2½×F for D in eq.2b: 2½×F - F = B which makes 1½×F = B


  

Solution

Substitute 3½×F for A and E, 1½×F for B, 0 for C, and 2½×F for D in eq.1: 3½×F + 1½×F + 0 + 2½×F + 3½×F + F = 24 which simplifies to 12×F = 24 Divide both sides of the above equation by 12: 12×F ÷ 12 = 24 ÷ 12 which means F = 2 making A = E = 3½×F = 3½ × 2 = 7 B = 1½×F = 1½ × 2 = 3 D = 2½×F = 2½ × 2 = 5 and ABCDEF = 730572