Puzzle for November 26, 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) F = B + C eq.3) D + E = A + B eq.4) A + D = B + C + E eq.5) B – C + D = A – B + F eq.6) A + D + E – F = C + F

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

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Hint #1


In eq.6, substitute B + C + E for A + D (from eq.4), and (B + C) for F (from eq.2): B + C + E + E – (B + C) = C + (B + C) which becomes B + C + 2×E – B – C = 2×C + B which becomes 2×E = 2×C + B Subtract 2×C from each side of the equation above: 2×E – 2×C = 2×C + B – 2×C which makes eq.6a) 2×E – 2×C = B


  

Hint #2


In eq.2, replace B with 2×E – 2×C (from eq.6a): F = 2×E – 2×C + C which becomes eq.2a) F = 2×E – C


  

Hint #3


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


  

Hint #4


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


  

Hint #5


Substitute E – 2×C for D – A (from eq.3a) into eq.5a: E – 2×C = 4×C – 2×E Add 2×C and 2×E to both sides of the equation above: E – 2×C + 2×C + 2×E = 4×C – 2×E + 2×C + 2×E which makes 3×E = 6×C Divide both sides of the above equation by 3: 3×E ÷ 3 = 6×C ÷ 3 which makes E = 2×C


  

Hint #6


Substitute (2×C) for E in eq.6a: 2×(2×C) – 2×C = B which becomes 4×C – 2×C = B which makes 2×C = B


  

Hint #7


Substitute (2×C) for E in eq.2a: F = 2×(2×C) – C which becomes F = 4×C – C which makes F = 3×C


  

Hint #8


Substitute 2×C for E in eq.3a: D – A = 2×C – 2×C which becomes D – A = 0 Add A to both sides of the above equation: D – A + A = 0 + A which makes D = A


  

Hint #9


Substitute A for D, and 2×C for B and E in eq.4: A + A = 2×C + C + 2×C which makes 2×A = 5×C Divide both sides of the above equation by 2: 2×A ÷ 2 = 5×C ÷ 2 which makes A = 2½×C and also makes D = A = 2½×C


  

Solution

Substitute 2½×C for A and D, 2×C for B and E, and 3×C for F in eq.1: 2½×C + 2×C + C + 2½×C + 2×C + 3×C = 26 which simplifies to 13×C = 26 Divide both sides of the above equation by 13: 13×C ÷ 13 = 26 ÷ 13 which means C = 2 making A = D = 2½×C = 2½ × 2 = 5 B = E = 2×C = 2 × 2 = 4 F = 3×C = 3 × 2 = 6 and ABCDEF = 542546