Puzzle for August 27, 2021  ( )

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

eq.1) A + B + C + D + E + F = 26 eq.2) B = C + F eq.3) D – B = A + B eq.4) E = A + B + C eq.5) D – F = B + C + F eq.6) C + E – F = B – C + D

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

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Hint #1


Add F to both sides of eq.5: D – F + F = B + C + F + F which becomes D = B + C + F + F In the above equation, replace B for C + F (from eq.2): D = B + B + F which becomes eq.5a) D = 2×B + F


  

Hint #2


In eq.3, replace D with 2×B + F (from eq.5a): 2×B + F – B = A + B which becomes B + F = A + B Subtract B from each side of the equation above: B + F – B = A + B – B which makes F = A


  

Hint #3


In eq.2, substitute A for F: eq.2a) B = C + A


  

Hint #4


eq.4 may be written as: E = C + A + B Substitute B for C + A (from eq.2a) into the above equation: E = B + B which makes eq.4a) E = 2×B


  

Hint #5


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


  

Hint #6


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


  

Hint #7


Substitute (C + A) for B (from eq.2a) in eq.6a: 2×(C + A) = C + 5×A which is equivalent to 2×C + 2×A = C + 5×A Subtract 2×A and C from each side of the equation above: 2×C + 2×A – 2×A – C = C + 5×A – 2×A – C which makes C = 3×A


  

Hint #8


Substitute 3×A for C in eq.2a: B = 3×A + A which makes B = 4×A


  

Hint #9


Substitute (3×A) for C in eq.5b: D = 2×(3×A) + 3×A which becomes D = 6×A + 3×A which makes D = 9×A


  

Hint #10


Substitute (4×A) for B in eq.4a: E = 2×(4×A) which makes E = 8×A


  

Solution

Substitute 4×A for B, 3×A for C, 9×A for D, 8×A for E, and A for F in eq.1: A + 4×A + 3×A + 9×A + 8×A + A = 26 which simplifies to 26×A = 26 Divide both sides of the above equation by 26: 26×A ÷ 26 = 26 ÷ 26 which means A = 1 making B = 4×A = 4 × 1 = 4 C = 3×A = 3 × 1 = 3 D = 9×A = 9 × 1 = 9 E = 8×A = 8 × 1 = 8 F = A = 1 and ABCDEF = 143981