Puzzle for February 26, 2019  ( )

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

eq.1) A + B + C + D + E + F = 17 eq.2) E + F = A - D eq.3) F = C + D + E eq.4) A - C = D - E eq.5)* C - B = B + (C ^ D)

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

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Hint #1


Add C + E to both sides of eq.4: A - C + C + E = D - E + C + E which becomes A + E = D + C which is equivalent to A + E = C + D In eq.3, replace C + D with A + E: F = A + E + E which becomes eq.3a) F = A + 2×E


  

Hint #2


In eq.2, replace F with A + 2×E (from eq.3a): E + A + 2×E = A - D Subtract A from each side of the equation above: E + A + 2×E - A = A - D - A which simplifies to 3×E = -D Add D to each side: 3×E + D = -D + D which becomes 3×E + D = 0 Since D and E must be one-digit non-negative integers, the above equation makes E = 0 and D = 0


  

Hint #3


In eq.2, substitute 0 for both E and D: 0 + F = A - 0 which makes F = A


  

Hint #4


In eq.3, substitute 0 for both D and E: F = C + 0 + 0 which means F = C


  

Hint #5


Substitute 0 for D in eq.5: C - B = B + (C ^ 0) Add B to each side of the above equation: C - B + B = B + (C ^ 0) + B which becomes C = 2×B + (C ^ 0) Since C ^ 0 = 1 for any positive number, then C = 2×B + 1 (implies C ≠ 0) which also means eq.5a) A = F = C = 2×B + 1


  

Solution

Substitute 2×B + 1 for A and C and F (from eq.5a), and 0 for D and E in eq.1: 2×B + 1 + B + 2×B + 1 + 0 + 0 + 2×B + 1 = 17 which simplifies to 7×B + 3 = 17 Subtract 3 from each side of the above equation: 7×B + 3 - 3 = 17 - 3 which means 7×B = 14 Divide both sides by 7: 7×B ÷ 7 = 14 ÷ 7 which makes B = 2 making A = F = C = 2×B + 1 = 2×2 + 1 = 5 (from eq.5a) and ABCDEF = 525005