Puzzle for December 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 = 25 eq.2) E - C = F - B + D eq.3) C + F = A + E - D eq.4) A + B = C - A + D eq.5) E + D = A + B + F - D eq.6) B + C - F = E + F - D

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

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Hint #1


Add C and B to both sides of eq.2: E - C + C + B = F - B + D + C + B which becomes E + B = F + D + C which may be written as eq.2a) B + E = C + D + F   Add D to both sides of eq.3: C + F + D = A + E - D + D which becomes C + F + D = A + E which may be written as eq.3a) C + D + F = A + E


  

Hint #2


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


  

Hint #3


In eq.4, replace B with A: A + A = C - A + D which becomes 2×A = C - A + D Add A to both sides of the above equation: 2×A + A = C - A + D + A which becomes eq.4a) 3×A = C + D


  

Hint #4


In eq.3a, substitute 3×A for C + D (from eq.4a): 3×A + F = A + E Subtract A from each side of the above equation: 3×A + F - A = A + E - A which becomes eq.2b) 2×A + F = E


  

Hint #5


Substitute 2×A + F for E (from eq.2b), and A for B in eq.5: 2×A + F + D = A + A + F - D which becomes 2×A + F + D = 2×A + F - D Subtract 2×A, F, and D from both sides of the equation above: 2×A + F + D - 2×A - F - D = 2×A + F - D - 2×A - F - D which makes 0 = -2×D which means 0 = D


  

Hint #6


Substitute 0 for D in eq.4a: 3×A = C + 0 which makes eq.4b) 3×A = C


  

Hint #7


Substitute A for B, 3×A for C, 2×A + F for E (from eq.2b), and 0 for D in eq.6: A + 3×A - F = 2×A + F + F - 0 which becomes 4×A - F = 2×A + 2×F In the above equation, add F to both sides, and subtract 2×A from both sides: 4×A - F + F - 2×A = 2×A + 2×F + F - 2×A which makes 2×A = 3×F Divide both sides by 2: 2×A ÷ 2 = 3×F ÷ 2 which makes A = 1½×F and also makes B = A = 1½×F


  

Hint #8


Substitute (1½×F) for A in eq.4b: 3×(1½×F) = C which makes 4½×F = C


  

Hint #9


Substitute (1½×F) for A in eq.2b: 2×(1½×F) + F = E which becomes 3×F + F = E which makes 4×F = E


  

Solution

Substitute 1½×F for A and B, 4½×F for C, 0 for D, and 4×F for E in eq.1: 1½×F + 1½×F + 4½×F + 0 + 4×F + F = 25 which simplifies to 12½×F = 25 Divide both sides of the above equation by 12½: 12½×F ÷ 12½ = 25 ÷ 12½ which means F = 2 making A = B = 1½×F = 1½ × 2 = 3 C = 4½×F = 4½ × 2 = 9 E = 4×F = 4 × 2 = 8 and ABCDEF = 339082