Puzzle for February 3, 2021  ( )

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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 = A + B eq.3) C + D + F = A + E eq.4) B + C = A + E + F eq.5) E + F = A + C eq.6) C = A + D

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

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Hint #1


In eq.4, replace E + F with A + C (from eq.5): B + C = A + A + C which becomes B + C = 2×A + C Subtract C from each side of the equation above: B + C – C = 2×A + C – C which makes B = 2×A


  

Hint #2


In eq.2, replace B with 2×A: E = A + 2×A which makes E = 3×A


  

Hint #3


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


  

Hint #4


eq.1 may be written as: A + B + E + C + D + F = 30 Substitute 2×A for B, 3×A for E, and 4×A for C + D + F (from eq.3a) in the above equation: A + 2×A + 3×A + 4×A = 30 which makes 10×A = 30 Divide both sides by 10: 10×A ÷ 10 = 30 ÷ 10 which means A = 3 making B = 2×A = 2 × 3 = 6 E = 3×A = 3 × 3 = 9


  

Hint #5


Substitute 3 for A, and 9 for E in eq.5: 9 + F = 3 + C Subtract 9 from each side of the above equation: 9 + F – 9 = 3 + C – 9 which becomes eq.5a) F = C – 6


  

Hint #6


Substitute 3 for A in eq.6: C = 3 + D Subtract 3 from each side of the equation above: C – 3 = 3 + D – 3 which becomes eq.6a) C – 3 = D


  

Solution

Substitute C – 3 for D (from eq.6a), C – 6 for F (from eq.5a), 3 for A, and 9 for E in eq.3: C + C – 3 + C – 6 = 3 + 9 which becomes 3×C – 9 = 12 Add 9 to both sides of the equation above: 3×C – 9 + 9 = 12 + 9 which makes 3×C = 21 Divide both sides of the equation above by 3: 3×C ÷ 3 = 21 ÷ 3 which means C = 7 making D = C – 3 = 7 – 3 = 4 (from eq.6a) F = C – 6 = 7 – 6 = 1 (from eq.5a) and ABCDEF = 367491