Puzzle for December 2, 2020  ( )

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

eq.1) A + B + C + D + E + F = 20 eq.2) C + D + F = A + B eq.3) B = E – D eq.4) C = E – B – F eq.5) E – A – D = A + D eq.6) A + D + F = E – A

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

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Hint #1


In eq.6, replace A + D with E – A – D (from eq.5): E – A – D + F = E – A Add A and D to both sides of the above equation: E – A – D + F + A + D = E – A + A + D which becomes E + F = E + D Subtract E from both sides: E + F – E = E + D – E which makes F = D


  

Hint #2


Add A and D to both sides of eq.5: E – A – D + A + D = A + D + A + D which becomes eq.5a) E = 2×A + 2×D   In eq.3, replace E with 2×A + 2×D (from eq.5a): B = 2×A + 2×D – D which becomes eq.3a) B = 2×A + D


  

Hint #3


In eq.4, substitute 2×A + 2×D for E (from eq.5a), 2×A + D for B (from eq.3a), and D for F: C = 2×A + 2×D – (2×A + D) – D which becomes C = 2×A + 2×D – 2×A – D – D which makes C = 0


  

Hint #4


Substitute 0 for C, D for F, and 2×A + D for B (from eq.3a) in eq.2: 0 + D + D = A + 2×A + D which becomes 2×D = 3×A + D Subtract D from both sides of the above equation: 2×D – D = 3×A + D – D which makes D = 3×A and also makes F = D = 3×A


  

Hint #5


Substitute 3×A for D in eq.3a: B = 2×A + 3×A which makes B = 5×A


  

Hint #6


Substitute (3×A) for D in eq.5a: E = 2×A + 2×(3×A) which becomes E = 2×A + 6×A which makes E = 8×A


  

Solution

Substitute 5×A for B, 0 for C, 3×A for D and F, and 8×A for E in eq.1: A + 5×A + 0 + 3×A + 8×A + 3×A = 20 which simplifies to 20×A = 20 Divide both sides of the equation above by 20: 20×A ÷ 20 = 20 ÷ 20 which means A = 1 making B = 5×A = 5 × 1 = 5 D = F = 3×A = 3 × 1 = 3 E = 8×A = 8 × 1 = 8 and ABCDEF = 150383