Puzzle for October 9, 2020  ( )

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

eq.1) A + B + C + D + E + F = 44 eq.2) B + D + F = C + E eq.3) A + B – C = D + E – F eq.4) F – D = A – F eq.5) D – C = B – D – E eq.6) B = average (A, C, E, F)

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

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Hint #1


Subtract D from both sides of eq.2: B + D + F – D = C + E – D which becomes eq.2a) B + F = C + E – D   Add F to both sides of eq.3: A + B – C + F = D + E – F + F which becomes eq.3a) A + B + F – C = D + E


  

Hint #2


In eq.3a, replace B + F with C + E – D (from eq.2a): A + C + E – D – C = D + E which becomes A + E – D = D + E In the above equation, subtract E from each side, and add D to each side: A + E – D – E + D = D + E – E + D which simplifies to A = 2×D


  

Hint #3


In eq.4, replace A with 2×D: F – D = 2×D – F Add D and F to both sides of the above equation: F – D + D + F = 2×D – F + D + F which becomes 2×F = 3×D Divide both sides by 2: 2×F ÷ 2 = 3×D ÷ 2 which makes F = 1½×D


  

Hint #4


eq.6 may be written as: B = (A + C + E + F) ÷ 4 Multiply both sides of the equation above by 4: 4 × B = 4 × (A + C + E + F) ÷ 4 which becomes eq.6a) 4×B = A + C + E + F


  

Hint #5


In eq.6a, substitute 2×D for A, and B + D + F for C + E (from eq.2): 4×B = 2×D + B + D + F + F which becomes 4×B = 3×D + B + 2×F Subtract B from each side of the equation above: 4×B – B = 3×D + B + 2×F – B which becomes eq.6b) 3×B = 3×D + 2×F


  

Hint #6


Substitute (1½×D) for F in eq.6b: 3×B = 3×D + 2×(1½×D) which becomes 3×B = 3×D + 3×D which makes 3×B = 6×D Divide both sides of the above equation by 3: 3×B ÷ 3 = 6×D ÷ 3 which makes B = 2×D


  

Hint #7


Add C and D to both sides of eq.5: D – C + C + D = B – D – E + C + D which becomes 2×D = B – E + C Subtract B from both sides of the equation above: 2×D – B = B – E + C – B which becomes eq.5a) 2×D – B = –E + C


  

Hint #8


Add the left and right sides of eq.5a to the left and right sides of eq.2, respectively: B + D + F + 2×D – B = C + E + (–E + C) which becomes F + 3×D = 2×C Substitute 1½×D for F in the above equation: 1½×D + 3×D = 2×C which becomes 4½×D = 2×C Divide both sides by 2: 4½×D ÷ 2 = 2×C ÷ 2 which makes 2¼×D = C


  

Hint #9


Substitute 2×D for B, and 2¼×D for C in eq.5a: 2×D – 2×D = –E + 2¼×D which becomes 0 = –E + 2¼×D Add E to both sides of the equation above: 0 + E = –E + 2¼×D + E which makes E = 2¼×D


  

Solution

Substitute 2×D for A and B, 2¼×D for C and E, and 1½×D for F in eq.1: 2×D + 2×D + 2¼×D + D + 2¼×D + 1½×D = 44 which simplifies to 11×D = 44 Divide both sides of the above equation by 11: 11×D ÷ 11 = 44 ÷ 11 which means D = 4 making A = B = 2×D = 2 × 4 = 8 C = E = 2¼×D = 2¼ × 4 = 9 F = 1½×D = 1½ × 4 = 6 and ABCDEF = 889496