Puzzle for September 26, 2020  ( )

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

eq.1) A + B + C + D + E + F = 38 eq.2) D + E – A = A + C + F eq.3) C + D – F = B – A eq.4) F – C = A eq.5) A + E = C + F – A eq.6) B + D – F = A + 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, substitute (F – C) for A (from eq.4): D + E – (F – C) = (F – C) + C + F which becomes D + E – F + C = 2×F Add F to both sides of the equation above: D + E – F + C + F = 2×F + F which becomes eq.2a) D + E + C = 3×F


  

Hint #2


In eq.6, replace A with F – C (from eq.4): B + D – F = F – C + F which becomes B + D – F = 2×F – C Add F and C to both sides of the equation above: B + D – F + F + C = 2×F – C + F + C which becomes eq.6a) B + D + C = 3×F


  

Hint #3


In eq.2a, replace 3×F with B + D + C from eq.6a: D + E + C = B + D + C Subtract D and C from both sides of the above equation: D + E + C – D – C = B + D + C – D – C which simplifies to E = B


  

Hint #4


In eq.3, substitute E for B: eq.3a) C + D – F = E – A


  

Hint #5


In eq.2, substitute C + D – F for E – A (from eq.3a): D + C + D – F = A + C + F which becomes C + 2×D – F = A + C + F In the above equation, subtract C from each side, and add F to each side: C + 2×D – F – C + F = A + C + F – C + F which becomes 2×D = A + 2×F Divide both sides by 2: 2×D ÷ 2 = (A + 2×F) ÷ 2 which becomes eq.2b) D = ½×A + F


  

Hint #6


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


  

Hint #7


Substitute ½×A + F for D (from eq.2b), 2×F – 3×A for E, and F – A for C (from eq.4a) in eq.2a: ½×A + F + 2×F – 3×A + F – A = 3×F which becomes 4×F – 3½×A = 3×F In the equation above, add 3½×A to both sides, and subtract 3×F from both sides: 4×F – 3½×A + 3½×A – 3×F = 3×F + 3½×A – 3×F which makes F = 3½×A


  

Hint #8


Substitute 3½×A for F in eq.5a: 3½×A – A = C which makes 2½×A = C


  

Hint #9


Substitute 3½×A for F in eq.2b: D = ½×A + 3½×A which makes D = 4×A


  

Hint #10


Substitute (3½×A) for F in eq.5a: E = 2×(3½×A) – 3×A which becomes E = 7×A – 3×A which makes E = 4×A and also makes B = E = 4×A


  

Solution

Substitute 4×A for B and D and E, 2½×A for C, and 3½×A for F in eq.1: A + 4×A + 2½×A + 4×A + 4×A + 3½×A = 38 which simplifies to 19×A = 38 Divide both sides of the equation above by 19: 19×A ÷ 19 = 38 ÷ 19 which means A = 2 making B = D = E = 4×A = 4 × 2 = 8 C = 2½×A = 2½ × 2 = 5 F = 3½×A = 3½ × 2 = 7 and ABCDEF = 285887