Puzzle for November 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 = 41 eq.2) D + E = A + C eq.3) A + D + E – F = C + F eq.4) E – C = C + D – E eq.5) C + D = A – D + F eq.6) B – (F – B – 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.3, replace D + E with A + C (from eq.2): A + A + C – F = C + F which becomes 2×A + C – F = C + F In the equation above, subtract C from both sides, and add F to both sides: 2×A + C – F – C + F = C + F – C + F which simplifies to 2×A = 2×F Divide both sides by 2: 2×A ÷ 2 = 2×F ÷ 2 which makes A = F


  

Hint #2


Add C and E to both sides of eq.4: E – C + C + E = C + D – E + C + E which becomes eq.4a) 2×E = 2×C + D   Multiply both sides of eq.2 by 2: 2×(D + E) = 2×(A + C) which is the same as eq.2a) 2×D + 2×E = 2×A + 2×C


  

Hint #3


In eq.2a, replace 2×E with 2×C + D (from eq.4a): 2×D + 2×C + D = 2×A + 2×C which becomes 3×D + 2×C = 2×A + 2×C Subtract 2×C from each side of the above equation: 3×D + 2×C – 2×C = 2×A + 2×C – 2×C which makes 3×D = 2×A Divide both sides by 2: 3×D ÷ 2 = 2×A ÷ 2 which makes 1½×D = A and also makes F = A = 1½×D


  

Hint #4


In eq.5, substitute 1½×D for A and F: C + D = 1½×D – D + 1½×D which becomes C + D = 2×D Subtract D from each side of the equation above: C + D – D = 2×D – D which makes eq.5a) C = D


  

Hint #5


Substitute D for C in eq.4a: 2×E = 2×D + D which makes 2×E = 3×D Divide both sides by 2: 2×E ÷ 2 = 3×D ÷ 2 which means E = 1½×D and means eq.4b) E = F = A = 1½×D


  

Hint #6


Substitute 1½×D for A and F, and D for C in eq.6: B – (1½×D – B – D – D) = 1½×D – B which is equivalent to B – 1½×D + B + D + D = 1½×D – B which becomes 2×B + ½×D = 1½×D – B In the equation above, subtract ½×D from both sides, and add B to both sides: 2×B + ½×D – ½×D + B = 1½×D – B – ½×D + B which makes 3×B = D and also makes C = D = 3×B


  

Hint #7


Substitute (3×B) for D in eq.4b: E = F = A = 1½×(3×B) which means E = F = A = 4½×B


  

Solution

Substitute 4½×B for A and E and F, and 3×B for C and D in eq.1: 4½×B + B + 3×B + 3×B + 4½×B + 4½×B = 41 which simplifies to 20½×B = 41 Divide both sides of the above equation by 20½: 20½×B ÷ 20½ = 41 ÷ 20½ which means B = 2 making A = E = F = 4½×B = 4½ × 2 = 9 C = D = 3×B = 3 × 2 = 6 and ABCDEF = 926699