Puzzle for December 16, 2020  ( )

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

eq.1) A + B + C + D + E + F = 18 eq.2) E = average (C, D, F) eq.3) A = average (B, C, D, E, F) eq.4) F = A×E eq.5)* A – E = F ^ D

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  "F ^ D" means "F raised to the power of D".

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Hint #1


eq.3 may be written as: A = (B + C + D + E + F) ÷ 5 Multiply both sides of the equation above by 5: 5 × A = 5 × (B + C + D + E + F) ÷ 5 which becomes eq.3a) 5×A = B + C + D + E + F


  

Hint #2


In eq.1, replace B + C + D + E + F with 5×A (from eq.3a): A + 5×A = 18 which makes 6×A = 18 Divide both sides of the above equation by 6: 6×A ÷ 6 = 18 ÷ 6 which makes A = 3


  

Hint #3


In eq.4, substitute 3 for A: F = 3×E


  

Hint #4


eq.2 may be written as: E = (C + D + F) ÷ 3 Multiply both sides of the equation above by 3: 3 × E = 3 × (C + D + F) ÷ 3 which becomes eq.2a) 3×E = C + D + F


  

Hint #5


In eq.2a, substitute F for 3×E: F = C + D + F Subtract F from both sides of the equation above: F – F = C + D + F – F which makes 0 = C + D Since C and D must be non-negative integers, the above equation makes: C = 0 and D = 0


  

Hint #6


Substitute 3 for A, and 0 for D in eq.5: 3 – E = F ^ 0 which becomes 3 – E = 1 (implies F ≠ 0) In the above equation, add E to both sides, and subtract 1 from both sides: 3 – E + E – 1 = 1 + E – 1 which makes 2 = E and also makes F = 3×E = 3×2 = 6


  

Solution

Substitute 3 for A, 0 for C and D, 2 for E, and 6 for F in eq.3a: 5×3 = B + 0 + 0 + 2 + 6 which becommes 15 = B + 8 Subtract 8 from each side of the equation above: 15 – 8 = B + 8 – 8 which makes 7 = B and ABCDEF = 370026