Puzzle for January 7, 2021  ( )

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

eq.1) A + B + C + D + E + F = 29 eq.2) D + F = B + C – D eq.3) E + F = A + C eq.4) C + D = A + B + E eq.5) B + D = A – D eq.6) D + E = A + B – D

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

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Hint #1


Add D to both sides of eq.5: B + D + D = A – D + D which becomes eq.5a) B + 2×D = A   Add D to both sides of eq.6: D + E + D = A + B – D + D which becomes eq.6a) 2×D + E = A + B


  

Hint #2


In eq.6a, replace A with B + 2×D (from eq.5a): 2×D + E = B + 2×D + B which becomes 2×D + E = 2×B + 2×D Subtract 2×D from both sides of the above equation: 2×D + E – 2×D = 2×B + 2×D – 2×D which makes eq.6b) E = 2×B


  

Hint #3


In eq.4, replace A with B + 2×D (from eq.5a), and E with 2×B: C + D = B + 2×D + B + 2×B which becomes C + D = 4×B + 2×D Subtract D from each side of the above equation: C + D – D = 4×B + 2×D – D which becomes eq.4a) C = 4×B + D


  

Hint #4


In eq.3, substitute 2×B for E, B + 2×D for A (from eq.5a), and 4×B + D for C (from eq.4a): 2×B + F = B + 2×D + 4×B + D which becomes 2×B + F = 5×B + 3×D Subtract 2×B from each side of the equation above: 2×B + F – 2×B = 5×B + 3×D – 2×B which becomes eq.3a) F = 3×B + 3×D


  

Hint #5


Substitute 3×B + 3×D for F (from eq.3a), and 4×B + D for C (from eq.4a) in eq.2: D + 3×B + 3×D = B + 4×B + D – D which becomes 3×B + 4×D = 5×B Subtract 3×B from both sides of the equation above: 3×B + 4×D – 3×B = 5×B – 3×B which makes 4×D = 2×B Divide both sides by 2: 4×D ÷ 2 = 2×B ÷ 2 which makes 2×D = B


  

Hint #6


Substitute (2×D) for B in eq.3a: F = 3×(2×D) + 3×D which becomes F = 6×D + 3×D which makes F = 9×D


  

Hint #7


Substitute (2×D) for B in eq.4a: C = 4×(2×D) + D which becomes C = 8×D + D which makes C = 9×D


  

Hint #8


Substitute (2×D) for B in eq.6b: E = 2×(2×D) which makes E = 4×D


  

Hint #9


Substitute 2×D for B in eq.5a: 2×D + 2×D = A which makes 4×D = A


  

Solution

Substitute 4×D for A and E, 2×D for B, and 9×D for C and F in eq.1: 4×D + 2×D + 9×D + D + 4×D + 9×D = 29 which simplifies to 29×D = 29 Divide both sides of the above equation by 29: 29×D ÷ 29 = 29 ÷ 29 which means D = 1 making A = E = 4×D = 4 × 1 = 4 B = 2×D = 2 × 1 = 2 C = F = 9×D = 9 × 1 = 9 and ABCDEF = 429149