Puzzle for December 13, 2022  ( )

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

eq.1) A + B + C + D + E + F = 30 eq.2) E = B + C eq.3) B = A + C eq.4) E = A + D eq.5) A + B = C + F - A eq.6) B - D + E = C + D - A

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

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Hint #1


In eq.5, replace B with A + C (from eq.3): A + A + C = C + F - A which becomes 2×A + C = C + F - A In the above equation, subtract C from both sides, and add A to both sides: 2×A + C - C + A = C + F - A - C + A which makes 3×A = F


  

Hint #2


In eq.6, replace B with A + C (from eq.3), and E with A + D (from eq.4): A + C - D + A + D = C + D - A which becomes 2×A + C = C + D - A In the above equation, subtract C from both sides, and add A to both sides: 2×A + C - C + A = C + D - A - C + A which makes 3×A = D


  

Hint #3


In eq.4, substitute 3×A for D: E = A + 3×A which becomes E = 4×A


  

Hint #4


Substitute 4×A for E, and A + C for B (from eq.3) in eq.2: 4×A = A + C + C which becomes 4×A = A + 2×C Subtract A from each side of the equation above: 4×A - A = A + 2×C - A which makes 3×A = 2×C Divide both sides by 2: 3×A ÷ 2 = 2×C ÷ 2 which makes 1½×A = C


  

Hint #5


Substitute 1½×A for C in eq.3: B = A + 1½×A which makes B = 2½×A


  

Solution

Substitute 2½×A for B, 1½×A for C, 3×A for D and F, and 4×A for E in eq.1: A + 2½×A + 1½×A + 3×A + 4×A + 3×A = 30 which simplifies to 15×A = 30 Divide both sides of the above equation by 15: 15×A ÷ 15 = 30 ÷ 15 which means A = 2 making B = 2½×A = 2½ × 2 = 5 C = 1½×A = 1½ × 2 = 3 D = F = 3×A = 3 × 2 = 6 E = 4×A = 4 × 2 = 8 and ABCDEF = 253686