If k is an integer, is 1/(2^(1-k))

Question

FRESH GMAT CLUB TESTS QUESTION

If k is an integer, is  \frac{1}{2^{1-k}} < \frac{1}{8}  ?

(1) k is a prime number
(2) k^2 + 1 is a prime number

M36-36

chetan2u · Answer

Since 2^{1-k} and 8 are positive, we can cross multiply to get 2^{1-k} > 8... 2^{1-k}>2^3 ......or 1-k>3....k<-2 (I) tells us the minimum value of k is 2, so never <-2.. Answer is always NO (II) k can be 4 or -4...yes or no A

Archit3110 · Answer

#1
k is a prime number
k will always be +ve so value of  \frac{1}{2^{1-k}} < \frac{1}{8}  
shall always be NO
sufficient
#2
k^2 + 1 is a prime number
here k can both be + or -ve
so we get both yes and no ; insufficient
IMO A

exc4libur · Answer

k=integer…\frac{1}{2^{1-k}} < \frac{1}{8}…2^{-(1-k)}<2^{-3}…-1+k<-3…Question:k<-2?

(1) k is a prime number   sufic.

k=prime…k≥2…Answer: no

(2) k^2 + 1 is a prime number   insufic.

k^2= …k^2+1= …k= …Answer:yes/no

Answer (A)

Bunuel · Answer

Official Solution:

If  k  is an integer, is  \frac{1}{2^{1-k}} < \frac{1}{8}  ? 
 
Is  \frac{1}{2^{1-k}} < \frac{1}{8}  
 
Cross-multiply:  8 < 2^{1-k}  
 
Is  2^3 < 2^{1-k} ? 
 
Is  3 < 1- k ? 
 
Is  k < -2 ? 
 
(1)  k  is a prime number
 
Only positive numbers can be primes (the least prime being 2), thus this statement implies that  k \geq 2 , which give a NO answer to the question whether  k < -2 . Sufficient,
 
(2)  k^2 + 1  is a prime number
 
Test numbers:
 
If  k=4 , then the answer is NO but if  k=-4 , then the answer is YES. Not sufficient.

Answer: A

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If k is an integer, is 1/(2^(1-k)) < 1/8 ?

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