Puzzle for March 30, 2022  ( )

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

eq.1) E = A + C eq.2) A – C = E – A eq.3) D + E – A = A – C + F eq.4) E = average (A, B, C, D) eq.5) C × F = B + D eq.6) E = (B + D – E) ÷ C

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 E with A + C (from eq.1): A – C = A + C – A which becomes A – C = C Add C to both sides of the above equation: A – C + C = C + C which makes A = 2×C


  

Hint #2


In eq.1, replace A with 2×C: E = 2×C + C which makes E = 3×C


  

Hint #3


In eq.3, substitute 3×C for E, and 2×C for A: D + 3×C – 2×C = 2×C – C + F which becomes D + C = C + F Subtract C from each side of the above equation: D + C – C = C + F – C which makes D = F


  

Hint #4


eq.4 may be written as: E = (A + B + C + D) ÷ 4 Multiply both sides of the above equation by 4: 4 × E = 4 × (A + B + C + D) ÷ 4 which becomes eq.4a) 4×E = A + B + C + D


  

Hint #5


Substitute (3×C) for E, and 2×C for A into eq.4a: 4×(3×C) = 2×C + B + C + D which becomes 12×C = 3×C + B + D Subtract 3×C from both sides of the above equation: 12×C – 3×C = 3×C + B + D – 3×C which becomes eq.4b) 9×C = B + D


  

Hint #6


Substitute 9×C for B + D (from eq.4b) in eq.5: C × F = 9×C Since C ≠ 0 (from eq.6), divide both sides of the above equation by C: C × F ÷ C = 9×C ÷ C which makes F = 9 and also makes D = F = 9


  

Hint #7


Substitute 3×C for E, and 9×C for B + D (from eq.4b) in eq.6: 3×C = (9×C – 3×C) ÷ C which becomes 3×C = (6×C) ÷ C which becomes 3×C = 6 Divide both sides of the above equation by 3: 3×C ÷ 3 = 6 ÷ 3 which makes C = 2 and makes A = 2×C = 2 × 2 = 4 E = 3×C = 3 × 2 = 6


  

Solution

Substitute 2 for C, and 9 for D in eq.4b: 9×2 = B + 9 which becomes 18 = B + 9 Subtract 9 from both sides of the equation above: 18 – 9 = B + 9 – 9 which makes 9 = B and makes ABCDEF = 492969