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If (sqrt(2) + 1)^{99} = N + r, where N is a positive integer and 0 < r  < 1, then find (N + r)r.

 Dec 18, 2019
 #1
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(Sqrt(2) +1)^99 =[78486117 9029326838 1800946701 0929219214 + 1.2741106665 9806646221 8468259296 5611703655..... E-38] x 1.2741106665 9806646221 8468259296 5611703655..... E-38 = 1

 

The answer = 1

 Dec 18, 2019
 #2
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It is very remarkable fact that: [Sqrt(2) + 1]^99 is very nearly a WHOLE integer !!!. Here it is calculated to 100 decimal places:

 

[sqrt(2) + 1]^99 =78486117902932683818009467010929219214.000000000000000000000000000000000000012741106665980664622184682592965611703655250957377780370113573509566385801635499911578533439100076941402033744061215557831686. When you multiply it by its vert small fraction part, it = 1

 Dec 18, 2019
 #3
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Beats me.    Any takers for this one?

 Dec 18, 2019

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