Puzzle for December 10, 2018  ( )

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

eq.1) A + B + C + D + E + F = 22 eq.2) A + E = F eq.3) B + D = E eq.4) A + C = F - C eq.5) B + C = D eq.6) A + B = C + D

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

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Hint #1


From eq.5, substitute B + C for D in eq.6: A + B = C + B + C Subtract B from each side: A + B - B = C + B + C - B which means A = 2×C


  

Hint #2


In eq.4, replace A with 2×C: 2×C + C = F - C Add C to each side of the equation above: 2×C + C + C = F - C + C which simplifies to 4×C = F


  

Hint #3


Substitute 2×C for A, and 4×C for F in eq.2: 2×C + E = 4×C Subtract 2×C from each side of the above equation: 2×C + E - 2×C = 4×C - 2×C which makes E = 2×C


  

Hint #4


Substitute 2×C for E in eq.3: B + D = 2×C Subtract D from each side of the above equation: B = 2×C - D In eq.5, replace B with 2×C - D: 2×C - D + C = D Add D to each side of the above equation: 2×C - D + C + D = D + D which simplifies to 3×C = 2×D Divide both sides by 2: 3×C ÷ 2 = 2×D ÷ 2 which means 1½×C = D


  

Hint #5


Substitute 1½×C for D in eq.5: B + C = 1½×C Subtract C from both sides of the above equation: B + C - C = 1½×C - C which means B = ½×C


  

Solution

Substitute 2×C for A and E, ½×C for B, 1½×C for D, and 4×C for F in eq.1: 2×C + ½×C + C + 1½×C + 2×C + 4×C = 22 which becomes 11×C = 22 Divide both sides of the above equation by 11: 11×C ÷ 11 = 22 ÷ 11 which makes C = 2 making A = E = 2×C = 2 × 2 = 4 B = ½×C = ½ × 2 = 1 D = 1½×C = 1½ × 2 = 3 F = 4×C = 4 × 2 = 8 and ABCDEF = 412348