Puzzle for March 20, 2020  ( )

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

eq.1) A + B + C + D + E + F = 27 eq.2) C + E = A + B eq.3) E + F = B – F eq.4) D + F = E eq.5) B = C + D eq.6) A + C + F = D – F

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

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Hint #1


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


  

Hint #2


In eq.4, replace E with A + D (from eq.2a): D + F = A + D Subtract D from both sides of the above equation: D + F – D = A + D – D which makes F = A


  

Hint #3


Add F to both sides of eq.6: A + C + F + F = D – F + F which becomes A + C + 2×F = D In the equation above, substitute A for F: A + C + 2×A = D which becomes eq.6a) 3×A + C = D


  

Hint #4


Substitute 3×A + C for D (from eq.6a) in eq.2a: E = A + 3×A + C which becomes eq.2b) E = 4×A + C


  

Hint #5


Substitute 4×A + C for E (from eq.2b), and A for F in eq.3: 4×A + C + A = B – A which becomes 5×A + C = B – A Add A to both sides of the equation above: 5×A + C + A = B – A + A which becomes eq.3a) 6×A + C = B


  

Hint #6


Substitute 6×A + C for B (from eq.3a), and 3×A + C for D (from eq.6a) in eq.5: 6×A + C = C + 3×A + C which becomes 6×A + C = 2×C + 3×A Subtract C and 3×A from both sides of the equation above: 6×A + C – C – 3×A = 2×C + 3×A – C – 3×A which makes 3×A = C


  

Hint #7


Substitute 3×A for C in eq.3a: 6×A + 3×A = B which makes 9×A = B


  

Hint #8


Substitute 3×A for C in eq.6a: 3×A + 3×A = D which means 6×A = D


  

Hint #9


Substitute 3×A for C in eq.2b: E = 4×A + 3×A which makes E = 7×A


  

Solution

Substitute 9×A for B, 3×A for C, 6×A for D, and 7×A for E, and A for F in eq.1: A + 9×A + 3×A + 6×A + 7×A + A = 27 which simplifies to 27×A = 27 Divide both sides of the equation above by 27: 27×A ÷ 27 = 27 ÷ 27 which means A = 1 making B = 9×A = 9 × 1 = 9 C = 3×A = 3 × 1 = 3 D = 6×A = 6 × 1 = 6 E = 7×A = 7 × 1 = 7 F = A = 1 and ABCDEF = 193671