Puzzle for March 11, 2020  ( )

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

eq.1) A + B + C + D + E + F = 16 eq.2) C = A + B eq.3) C + D = B + E eq.4) B = E + F eq.5) D + E = C eq.6) E = D + 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 D + E (from eq.5): D + E + D = B + E Subtract E from each side of the above equation: D + E + D – E = B + E – E which makes 2×D = B


  

Hint #2


In eq.5, replace E with D + F (from eq.6): D + D + F = C which becomes eq.5a) 2×D + F = C


  

Hint #3


In eq.3, substitute 2×D + F for C (from eq.5a): 2×D + F + D = B + E which becomes eq.3a) 3×D + F = B + E


  

Hint #4


Substitute E + F for B (from eq.4) in eq.3a: 3×D + F = E + F + E which becomes 3×D + F = 2×E + F Subtract F from both sides of the equation above: 3×D + F – F = 2×E + F – F which means 3×D = 2×E Divide both sides by 2: 3×D ÷ 2 = 2×E ÷ 2 which makes 1½×D = E


  

Hint #5


Substitute 1½×D for E in eq.5: D + 1½×D = C which makes 2½×D = C


  

Hint #6


Substitute 1½×D for E in eq.6: 1½×D = D + F Subtract D from both sides of the equation above: 1½×D – D = D + F – D which makes ½×D = F


  

Hint #7


Substitute 2½×D for C, and 2×D for B in eq.2: 2½×D = A + 2×D Subtract 2×D from each side of the equation above: 2½×D – 2×D = A + 2×D – 2×D which makes ½×D = A


  

Solution

Substitute ½×D for A and F, 2×D for B, 2½×D for C, and 1½×D for E in eq.1: ½×D + 2×D + 2½×D + D + 1½×D + ½×D = 16 which simplifies to 8×D = 16 Divide both sides of the equation above by 8: 8×D ÷ 8 = 16 ÷ 8 which means D = 2 making A = F = ½×D = ½ × 2 = 1 B = 2×D = 2 × 2 = 4 C = 2½×D = 2½ × 2 = 5 E = 1½×D = 1½ × 2 = 3 and ABCDEF = 145231