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Bunuel
If k is an integer and \(0.01<\frac{1}{3^k}<0.1\), then what is the product of all possible value of k ?

A. -12
B. 0
C. 6
D. 12
E. 24
Given: \(\frac{1}{100}<\frac{1}{3^k}<\frac{1}{10}\)
Taking Reciprocal reverses the greater than/less than symbol:
\(100>3^k>10\)
So \(3^2 = 9; 3^3 = 27; 3^4 = 81; 3^5 = 243.\)
Out of these, only \( 3^3 = 27; 3^4 = 81\) lie between 10 and 100.
So k can only be 3 and 4.
3*4 = 12.
So the right answer is D.
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Bunuel
If k is an integer and \(0.01<\frac{1}{3^k}<0.1\), then what is the product of all possible value of k ?

A. -12
B. 0
C. 6
D. 12
E. 24

Another simple way to solve this for anyone who has trouble with decimals and fractions is:

\(0.01<\frac{1}{3^k}<0.1\)

Immediately try to get rid of the decimal by multiplying by 100.
You get:

\(1<\frac{100}{3^k}<10\)
So k has to be positive

if k = 1 then \(\frac{100}{3^1} = 33\) (NO)

if k = 2 then \(\frac{100}{3^2} = 11 \) (NO)

if k = 3 then \(\frac{100}{3^3} = 3. \) (YES)

if k = 4 then \(\frac{100}{3^4} = 1. \) (YES)

Only k = 3 and k = 4 is possible. So 3*4 = 12
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Hi
Since K is integer thus it can be +ve and -Ve both. In that Case K can be 3,4 and -4 as well so the desired answer should be -48.

Can you explain where am I missing?
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deepak.brijendra
Hi
Since K is integer thus it can be +ve and -Ve both. In that Case K can be 3,4 and -4 as well so the desired answer should be -48.

Can you explain where am I missing?

k = -4 does not satisfy 0.01 < 1/3^k < 0.1 because \(\frac{1}{3^{-4} }=3^4 = 81\).
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