Puzzle for February 23, 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) C = A + B eq.3) B + D = C + E eq.4) D + E = A + F eq.5) E - F = B - E eq.6) C = average (B, D, E, F)

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

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Hint #1


In eq.3, replace C with A + B (from eq.2): B + D = A + B + E Subtract B from both sides of the above equation: B + D - B = A + B + E - B which becomes eq.3a) D = A + E


  

Hint #2


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


  

Hint #3


In eq.5, substitute 2×E for F: E - 2×E = B - E which becomes -E = B - E Add E to both sides of the equation above: -E + E = B - E + E which makes 0 = B


  

Hint #4


Substitute 0 for B in eq.2: C = A + 0 which makes C = A


  

Hint #5


eq.6 may be written as: C = (B + D + E + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × C = 4 × (B + D + E + F) ÷ 4 which becomes eq.6a) 4×C = B + D + E + F


  

Hint #6


In eq.6a, substitute A for C, 0 for B, and 2×E for F: 4×A = 0 + D + E + 2×E which becomes eq.6b) 4×A = D + 3×E


  

Hint #7


Substitute A + E for D (from eq.3a) in eq.6b: 4×A = A + E + 3×E which becomes 4×A = A + 4×E Subtract A from each side of the above equation: 4×A - A = A + 4×E - A which makes 3×A = 4×E Divide both sides by 4: 3×A ÷ 4 = 4×E ÷ 4 which makes ¾×A = E


  

Hint #8


Substitute ¾×A for E in eq.3a: D = A + ¾×A which makes D = 1¾×A


  

Hint #9


Substitute (¾×A) for E in eq.4a: 2×(¾×A) = F which makes 1½×A = F


  

Solution

Substitute A for C, 0 for B, 1¾×A for D, ¾×A for E, and 1½×A for F in eq.1: A + 0 + A + 1¾×A + ¾×A + 1½×A = 24 which simplifies to 6×A = 24 Divide both sides of the above equation by 6: 6×A ÷ 6 = 24 ÷ 6 which means A = 4 making C = A = 4 D = 1¾×A = 1¾ × 4 = 7 E = ¾×A = ¾ × 4 = 3 F = 1½×A = 1½ × 4 = 6 and ABCDEF = 404736