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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 03:56
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3^-n must lie between 30 & 3000
3^3=27...so first -n=4
Then 243, 729, 2187.. next will be in 6K

so, -n= 4,5,6,7
or n= -7,-6,-5,-4
Median = avg of -6 & -5 = -5.5

Ans D
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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 04:00
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We are given 3000> 3^-n>30.
So we should check what powers of 3 qualify so that the value of 1/3^n or 3^-n fall within the range of 30 and 3000.

3^3= 27 (not possible since <30)
3^4= 81 (possible)
3^5= 243 (possible)
3^6 = 729 (possible)
3^7= 2187 (possible) (note that it should be clear by now without further calculation that anything greater than this will fall out of the end limit 3000.)

so possible values of -n are: -4, -5, -6, -7
median would be (-5)+(-6)/2 = -11/2

Hence, option D.
Bunuel
If n is an integer and 3000 > 1/3^n > 30, what is the median of all possible values of n?

A. -7
B. -13/2
C. -6
D. -11/2
E. -5


 


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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 04:48
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Here, 3000 >3^-n>30

so minimum is more than 30, that's minimum is 3^4

and maximum is 3000, i.e 3^4, 3^5,3^6,3^7, not 3^8 as 3^8=6561.

so the range of n= {-4,-5,-6,-7},
mean = (-6+(-5))/2
= -11/2

hence option D is correct
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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 05:11
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Bunuel
If n is an integer and 3000 > 1/3^n > 30, what is the median of all possible values of n?

A. -7
B. -13/2
C. -6
D. -11/2
E. -5


 


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The value of n should be such that middle term is <3000 but >30.
i.e., in place of n- -4,-5,-6 & -7 (3^3=27 < 30 & 3^8= 6561 > 3000).
Therefore, median of even numbers- avg of two middle nos. i.e. -5 & -6. -5.5 Option D
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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 05:13
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3000 > 3^(-n) >30
1000 > 3^ -(n +1) >10
possible values of -(n +1) = 3,4,5,6
so n = { -4, -5, -6, -7}

so median. = (-5 -6)/ 2 = -11/2
D
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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 05:30
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n can take -4,-5,-6,-7 that satisfy the condition
so median is -5.5 (-5-6/2) sum of 2nd and 3rd term
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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 05:33
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Bunuel
If n is an integer and 3000 > 1/3^n > 30, what is the median of all possible values of n?

A. -7
B. -13/2
C. -6
D. -11/2
E. -5


 


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Given that (1/3)^n is less than 3000 and more than 30
We know that 1/3 for any positive n, will be less than 30 and will keep on diminishing to 0 as we go higher
However, for every negative value of n from -1 to - infinity, the value will keep on increasing

For instance, (1/3)^-1 = 3, (1/3)^-2 = 9 and so on

We know that the minimum value that makes the value of 1/3 greater than 30 is -4
Similarly, the greatest value of n such that 1/3 is less than 3000 is -7

This gives us the set as {-4,-5,-6,-7}
The median is the average of the 2nd and 3rd term, giving us D) -11/2
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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 05:33
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3*10^3> (1/3)^n > 30 . In this format n has to be negative . (1/3)^-3= 27 , when n=-4 then its 81 when n=5 then its 243 and when n=6 it is 729 when n=-7 then its 2187 stop here

n ranges from -3, -4,-5,-6,-7 so median value is -5 ans is E
BunuelIf n is an integer and 3000 > 1/3^n > 30, what is the median of all possible values of n?

A. -7
B. -13/2
C. -6
D. -11/2
E. -5


 


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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 05:44
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30 < 3^(-n) < 3000

3^3 = 27
3^4 = 81
3^7 = 2187
3^8 = 6561

(-n) can be: 4, 5, 6, 7
n can be: -7, -6, -5, -4

median = (-6-5)/2 = -11/2

IMO D
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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 05:47
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Rewriting question stem as 3000 > 3^(-n) > 30
So n<0 for 3^(-n) > 0
now, 3^3 = 27 3^7=2187 & 3^8 = 6561
So power > 3 and power < 8
i.e. n<-3 and n>-8 (because n<0)
So all possible values of n are -4,-5,-6,-7
And median = (-5 + -6)/2 = -11/2

Option D
Bunuel
If n is an integer and 3000 > 1/3^n > 30, what is the median of all possible values of n?

A. -7
B. -13/2
C. -6
D. -11/2
E. -5


 


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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 05:57
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We can write this as,
30 < 3^-n < 3000
Now, 3^3 = 27, 3^4 = 81..., 3^7 = 2187 and 3^8 = 6561
So, we can have 3^4 <= 3^-n <= 3^7
Hence, 4<= -n <= 7
Multiplying by -1, it becomes -7 <=n <= -4

integer can be -> -7, -6, -5, -4
median = (-6 + (-5))/2 = -11/2
Answer D is correct
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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 05:59
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3000 > 1/(3^n) > 30
30 < 3^(-n) < 3000

3^3=27 and 3^4=81 -> -n >= 4
3^7 = 2187 and 3^8 = 6561 -> -n <= 7

4 <= -n <= 7
-7 <= n <= -4

values: -7,-6,-5,-4 whose median is (-6-5)/2 = -11/2

The answer is D
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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 06:04
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We are given the inequality

3000 > 3^-n > 30

Now we need to check the powers of 3 which fall between these values
We see that,
3^4 = 81 and 3^7 = 2187 are the 2 extreme ends of power falling between 30 and 3000

So the possible values of -n are 4,5,6 and 7

=> 4 <= -n <= 7
=> -4 >= n >= -7

=> Valid values are -7, -6, -5, -4
=> Median = (-6 + (-5)) / 2 = -5.5


D. -11/2
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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 06:09
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I approached this question by trying to first simplify the equation.
Assuming we could reach the edges of the inequality, what would n be?

First, I decided to break the limits to their prime factors: 3000 = 3 * 1000= 3 * 10 * 100 = 3 * 10 * 10 * 10 = 3 * 10^3= 3 *(2 * 5) ^3 = 2^3 * (3) * 5^3

The other end is 3 * 10 = 2 * 3 * 5

This didn't help.

So, next I decided to divide the inequality by 30 and got 100 > (1/30) * (1/3^n) > 1
and simplifying 100 > 1/3 * 1/3^n * 1/10 > 1
(I should have divided just by 3)
I got 100 > (1/10) * 1/3^(n+1) > 1
When is the central expression equal the right limit?
(1/10) * 1/3^(n+1) = 1
1/3^(n+1) = 10
3 in the second power is close to 10
so, 1/3 in the -2nd power is close to 10. Exactly it's -7/3 power
n + 1 = -7/3
n = -10/3 is to one-third less than minus two
meaning -3 if I ceil it to an integer.

Taking on the left limit-
(1/10) * 1/3^(n+1) = 100
1/3^(n+1)=1000
n + 1 is a little more than minus six
So, n is a little more than minus seven.
Ceiling, we get -7.

n is between -
listing all the integers in this range, we get:
-7, -6, -5, -4, -3
And -5 is my answer.

So Answer choice E.
Bunuel
If n is an integer and 3000 > 1/3^n > 30, what is the median of all possible values of n?

A. -7
B. -13/2
C. -6
D. -11/2
E. -5


 


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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 06:17
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IMO : D
since 30<1/3^n <3000
we know that n is going to be negative
hence,
30<3^k<3000
here we know that
3 power 3 = 27 hence it will start from k = 4 and also since 3 ^7 2187
therefore possible values of n are == -4, -5, -6, -7
hence median = (-5+-6)/2== -11/2
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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 06:25
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Bunuel
If n is an integer and 3000 > 1/3^n > 30, what is the median of all possible values of n?

A. -7
B. -13/2
C. -6
D. -11/2
E. -5


 


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File comment: Lets simplify the equation by dividing it by 3

we would get

1000> 3^ (-n-1) > 10
For this eqn to make sence n has to be negative and it could be negative as n is an integer .

now lets say n = -3 so the 3^(-n-1) becomes 9 whoch is not greater than 10

so min value of n is -4

lets find the max value ,

3^6 is 729 and 3^7 is 2187 (greater than 1000) so,

-n-1 = 6 so n = -7 (max)

now lets write all possible values of n

-4, -5,-6, -7

median is -11/2
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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 06:25
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so say 1/3^n, can be >1 if n is -ve
30<3^-n<3000

and n that satisfies min=4 (3^4=81)
and max=7(3^7=2187)

hence, -4,-5,-6,-7 are the values
median will be (-5-6)/2=-11/2

Ans. D
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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 06:35
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(1/3)^n = 3^(-n)

So, 3000 > 3^(-n) > 30
3^3=27, 3^4=81
So 3^(-n) > 30, -n>=4

Also, 3^7 = 2187, 3^8=6561
So 3^(-n) < 3000 implies -n <= 7

So possible values of -n are 4,5,6,7
Or, possible values of n are -7, -6, -5, -4
The median value is (-6+-5)/2 = -11/2

Bunuel
If n is an integer and 3000 > 1/3^n > 30, what is the median of all possible values of n?

A. -7
B. -13/2
C. -6
D. -11/2
E. -5


 


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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 06:49
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If n is an integer and 3000 > 1/3^n > 30, what is the median of all possible values of n?

A. -7
B. -13/2
C. -6
D. -11/2
E. -5

n definitely less than 0
Say k = -n

So 3000 > 3^k > 30
If k is integer, then below can be only possible values of k
{4, 5, 6, 7}

Then n = {-7, -6, -5, -4}
Median = (-6-5)/2 = -11/2 ..Ans.(D)
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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n 09 Jul 2026, 07:08
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since we have \(\frac{1}{3^n}\) n must be a negative integer in order for the overall value to be greater than 1.

thus let us now check for \(3000 > 3^k > 30\)
since \(3^3 = 27\) and \(3^7 = 2187\) -> k must be in between 4 and 7 (inclusive)
therefore possible values of k are 4, 5, 6 and 7 which implies possible values of n are -4, -5, -6, -7
median is \(\frac{[(-5) + (-6)]}{2} = \frac{-11}{2}\)
Bunuel
If n is an integer and 3000 > 1/3^n > 30, what is the median of all possible values of n?

A. -7
B. -13/2
C. -6
D. -11/2
E. -5


 


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