Puzzle for January 14, 2023  ( )

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

eq.1) A + B + C + D + E + F = 29 eq.2) C + F = A + E eq.3) B + C = A + F eq.4) B + E = A - B - E + F eq.5) average (A, B, E) = D eq.6) average (B, C, E) = average (D, F)

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

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Hint #1


Subtract the left and right sides of eq.2 from the left and right sides of eq.3, respectively: B + C - (C + F) = A + F - (A + E) which becomes B + C - C - F = A + F - A - E which becomes B - F = F - E Add F to both sides of the above equation: B - F + F = F - E + F which becomes eq.3a) B = 2×F - E


  

Hint #2


eq.4 may be written as: B + E = A + F - B - E In the equation above, replace A + F with B + C (from eq.3): B + E = B + C - B - E which becomes B + E = C - E Add E to both sides: B + E + E = C - E + E which becomes eq.4a) B + 2×E = C


  

Hint #3


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


  

Hint #4


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


  

Hint #5


eq.5 may be written as: (A + B + E) ÷ 3 = D Multiply both sides of the above equation by 3: 3 × (A + B + E) ÷ 3 = 3 × D which becomes eq.5a) A + B + E = 3×D


  

Hint #6


Substitute 3×F for A, and 2×F - E for B (from eq.3a) in eq.5a: 3×F + 2×F - E + E = 3×D which becomes 5×F = 3×D Divide both sides of the above equation by 3: 5×F ÷ 3 = 3×D ÷ 3 which makes 1⅔×F = D


  

Hint #7


eq.6 may be written as: (B + C + E) ÷ 3 = (D + F) ÷ 2 Multiply both sides of the above equation by 6: 6 × (B + C + E) ÷ 3 = 6 × (D + F) ÷ 2 which becomes eq.6a) 2×B + 2×C + 2×E = 3×D + 3×F


  

Hint #8


In eq.6a, substitute (2×F - E) for B (from eq.3a), (2×F + E) for C (from eq.4b), and (1⅔×F) for D: 2×(2×F - E) + 2×(2×F + E) + 2×E = 3×(1⅔×F) + 3×F which becomes 4×F - 2×E + 4×F + 2×E + 2×E = 5×F + 3×F which becomes 8×F + 2×E = 8×F Subtract 8×F from each side of the equation above: 8×F + 2×E - 8×F = 8×F - 8×F which makes 2×E = 0 which means E = 0


  

Hint #9


Substitute 0 for E in eq.3a: B = 2×F - 0 which makes B = 2×F


  

Hint #10


Substitute 0 for E in eq.4b: 2×F + 0 = C which makes 2×F = C


  

Solution

Substitute 3×F for A, 2×F for B and C, 1⅔×F for D, and 0 for E in eq.1: 3×F + 2×F + 2×F + 1⅔×F + 0 + F = 29 which simplifies to 9⅔×F = 29 Divide both sides of the above equation by 9⅔: 9⅔×F ÷ 9⅔ = 29 ÷ 9⅔ which means F = 3 making A = 3×F = 3 × 3 = 9 B = C = 2×F = 2 × 3 = 6 D = 1⅔×F = 1⅔ × 3 = 5 and ABCDEF = 966503