Puzzle for August 24, 2020  ( )

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

eq.1) A + B + C + D + E + F = 31 eq.2) C + F = D eq.3) A = C + E eq.4) D – E = A + E eq.5) B + F = C + D eq.6) E + F = A + C

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

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Hint #1


In eq.6, replace A with C + E (from eq.3): E + F = C + E + C which becomes E + F = 2×C + E Subtract E from each side of the equation above: E + F – E = 2×C + E – E which makes F = 2×C


  

Hint #2


In eq.2, replace F with 2×C: C + 2×C = D which makes 3×C = D


  

Hint #3


In eq.5, replace D with C + F (from eq.2): B + F = C + C + F which becomes B + F = 2×C + F Subtract F from both sides of the above equation: B + F – F = 2×C + F – F which makes B = 2×C


  

Hint #4


Substitute 3×C for D, and C + E for A (from eq.3) in eq.4: 3×C – E = C + E + E which becomes 3×C – E = C + 2×E In the above equation, add E to both sides, and subtract C from both sides: 3×C – E + E – C = C + 2×E + E – C which makes 2×C = 3×E Divide both sides by 3: 2×C ÷ 3 = 3×E ÷ 3 which makes ⅔×C = E


  

Hint #5


Substitute ⅔×C for E in eq.3: A = C + ⅔×C which makes A = 1⅔×C


  

Solution

Substitute 1⅔×C for A, 2×C for B and F, 3×C for D, and ⅔×C for E in eq.1: 1⅔×C + 2×C + C + 3×C + ⅔×C + 2×C = 31 which simplifies to 10⅓×C = 31 Divide both sides of the equation above by 10⅓: 10⅓×C ÷ 10⅓ = 31 ÷ 10⅓ which means C = 3 making A = 1⅔×C = 1⅔ × 3 = 5 B = F = 2×C = 2 × 3 = 6 D = 3×C = 3 × 3 = 9 E = ⅔×C = ⅔ × 3 = 2 and ABCDEF = 563926