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Find the number of positive integers  n  that satisfy 

 May 16, 2019
 #1
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2^200  < n^100 < ( 130n)^50

 

Note that we can write

 

2^200  =  (2^2)^100  = 4^100  = (4^2)^50  = (16)^50

 

And  we can write   

 

n^100   =  (n^2) ^50

 

So  we have that

 

(16)^50 < (n^2)^50 < (130n)^50        which implies that

 

16 < n^2       so    n  > 4     and since we require an integer for n,  the solution to this part is  [ 5, inf)

 

And

 

n^2 < 130n

n^2 - 130n < 0

n(n - 130)< 0

This will be true   on the interval   ( 0, 130)  so....the interval for this part is  [ 1, 129 ]

 

So.....the most restrictive interval  that solves this is

 

[5, 129]

 

So   the number of positive integers is just    129 - 5 + 1   =   125  positive integers

 

 

 

cool cool cool

 May 16, 2019

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