Puzzle for April 22, 2019  ( )

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

eq.1) A + B + C + D + E + F = 20 eq.2) D = B + C eq.3) A + C + D = E + F eq.4) E - C = A + C eq.5) B + D = C + E + F eq.6) C + D = A + B

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 E + F with A + C + D (from eq.3): B + D = C + A + C + D Subtract D from both sides of the equation above: B + D - D = C + A + C + D - D which becomes eq.5a) B = A + 2×C


  

Hint #2


Add C to both sides of eq.4: E - C + C = A + C + C which becomes E = A + 2×C In eq.5a, replace A + 2×C with E: B = E


  

Hint #3


In eq.2, substitute E for B: D = E + C Substitute E + C for D, and E for B in eq.6: C + E + C = A + E Subtract E from both sides of the above equation: C + E + C - E = A + E - E which makes 2×C = A


  

Hint #4


Substitute 2×C for A in eq.5a: B = 2×C + 2×C which makes B = 4×C and also makes E = B = 4×C


  

Hint #5


Substitute 4×C for B in eq.2: D = 4×C + C which means D = 5×C


  

Hint #6


Substitute 5×C for D, and B for E in eq.5: B + 5×C = C + B + F Subtract both B and C from each side: B + 5×C - B - C = C + B + F - B - C which makes 4×C = F


  

Solution

Substitute 2×C for A, 4×C for B and E and F, and 5×C for D in eq.1: 2×C + 4×C + C + 5×C + 4×C + 4×C = 20 which becomes 20×C = 20 Divide both sides of the equation above by 20: 20×C ÷ 20 = 20 ÷ 20 which means C = 1 making A = 2×C = 2 × 1 = 2 B = E = F = 4×C = 4 × 1 = 4 D = 5×C = 5 × 1 = 5 and ABCDEF = 241544