Puzzle for April 2, 2020  ( )

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

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 A with E – D (from eq.3): E – D + B + C + D = E + F which becomes E + B + C = E + F Subtract E from both sides of the equation above: E + B + C – E = E + F – E which becomes eq.4a) B + C = F


  

Hint #2


In eq.4a, replace B with C – F (from eq.2): C – F + C = F which becomes 2×C – F = F Add F to both sides of the above equation: 2×C – F + F = F + F which means 2×C = 2×F Divide both sides by 2: 2×C ÷ 2 = 2×F ÷ 2 which makes C = F


  

Hint #3


In eq.2, substitute C for F: B = C – C which means B = 0


  

Hint #4


Add E to both sides of eq.6: F – E + E = A + B + E + E which becomes F = A + B + 2×E Substitute C for F, E – D for A (from eq.3), and 0 for B in the above equation: C = E – D + 0 + 2×E which becomes eq.6a) C = 3×E – D


  

Hint #5


Add D to both sides of eq.5: C – D + D = D + E + D which becomes C = 2×D + E Substitute 2×D + E for C in eq.6a: 2×D + E = 3×E – D In the equation above, subtract E from each side, and add D to each side: 2×D + E – E + D = 3×E – D – E + D which means 3×D = 2×E Divide both sides by 2: 3×D ÷ 2 = 2×E ÷ 2 which makes 1½×D = E


  

Hint #6


Substitute (1½×D) for E in eq.6a: C = 3×(1½×D) – D which becomes C = 4½×D – D which makes C = 3½×D and also makes F = C = 3½×D


  

Hint #7


Substitute 1½×D for E in eq.3: 1½×D – D = A which makes ½×D = A


  

Solution

Substitute ½×D for A, 0 for B, 3½×D for C and F, and 1½×D for E in eq.1: ½×D + 0 + 3½×D + D + 1½×D + 3½×D = 20 which simplifies to 10×D = 20 Divide both sides of the equation above by 10: 10×D ÷ 10 = 20 ÷ 10 which means D = 2 making A = ½×D = ½ × 2 = 1 C = F = 3½×D = 3½ × 2 = 7 E = 1½×D = 1½ × 2 = 3 and ABCDEF = 107237