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Question Number 134689 by benjo_mathlover last updated on 06/Mar/21

 ((2^(289) +1)/(2^(17) +1)) = 2^a_1   + 2^a_2   + 2^a_3   + ... + 2^a_k    Find the value of k.

2289+1217+1=2a1+2a2+2a3+...+2akFindthevalueofk.

Answered by EDWIN88 last updated on 06/Mar/21

Let 2^(17)  = y  ⇒ ((y^(17) +1)/(y+1)) = y^(16) −y^(15) +y^(14) −y^(13) +y^(12) −y^(11) +...+1   = y^(15) (y−1)+y^(13) (y−1)+y^(11) (y−1)+...+y(y−1) +1   = (y^(15) +y^(13) +y^(11) +y^9 +...+y)(y−1)+1   = (2^(255) +2^(221) +2^(187) +...+2^(17) )(2^(17) −1)+1   = (2^(255) +2^(221) +2^(187) +...+2^(17) )_(8) (2^(16) +2^(15) +2^(14) +...+2+1)_(17) +1  so we get k = 8×17+1 = 137

Let217=yy17+1y+1=y16y15+y14y13+y12y11+...+1=y15(y1)+y13(y1)+y11(y1)+...+y(y1)+1=(y15+y13+y11+y9+...+y)(y1)+1=(2255+2221+2187+...+217)(2171)+1=(2255+2221+2187+...+217)(216+215+214+...+2+1)178+1sowegetk=8×17+1=137

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