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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n

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harshitaa45

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08 Jul 2026, 09:23

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n is negative then only can this range work
3000>1/3^n>30
LHS: 3000>3^-n
as n is negative -n would be positive : maximum value of -n to satisfy it is 7 as 3^8=8001 and 3^7=2187.
-n=<7.
RHS:
3^-n>30 smallest value of -n is 4 as 3^3=27 and 3^4 is 81.
-n>=4
,-7->=-n>=4 => -7<=n<=-4 possible values of n : -7,-6,-5 and -4 so the median is -5.5=-11/2

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08 Jul 2026, 09:25

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The correct answer is -11/2. D.

We can rewrite the equation as 30 < 3^-n < 3000.

We need powers of 3 that fall in the range of 30 and 3000.
3^1=3
3^2=9
3^3= 27(<30)
3^4= 81
3^5= 243
3^6= 729
3^7= 2187
3^8= 6561 (>3000, so disqualified)

So the value of n that fits the equation would be
n=4,5,6,7
-n= -4,-5,-6,-7

the median of these values would be [(-5)+(-6)]/2
=-11/5

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The answer is (D), i.e., -(11/2)
1/3^n = 3^(-n)
Since 3,000 > 3^(-n) > 30,
if n = 3, 3^3 = 27
if n = 4, 3^4 = 81
if n = 5, 3^5 = 243
if n = 6, 3^6 = 729
if n = 7, 3^7 = 2,187
if n = 8, 3^8 = 6,561

Therefore, n = -4,-5,-6,-7 (27 is lower than 30 and 6,561 is greater than 3,000; hence, -3 and -8 are out).
The median will be = -(5+6)/2 = -11/2

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IMO D
Since n is an integer, clearly n should be negative
for n=-4, 1/3^n = 81 which is greater than 30
& for n= -7, 1/3^n = 2187, which is less than 3000
So median of -4, -5,-6, -7 = (-5-6)/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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08 Jul 2026, 09:35

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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?

3000 > 1/3^n > 30
3^3=27, 3^4=81.....3^7=2187, 3^8=6561.
value of n must be -4,-5,-6,-7

Median of all possible values of n= (-5+(-6))/2= -11/2

D

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First, note that:
1/(3^n) = 3^(-n)

=> 3000 > 3^(-n) > 30

Minimum (-n) to be greater than 30 = 4 which means n starts from -4
Maximum (-n) to be less than 3000 = 7 which means n ends at -7

n = {-4,-5,-6,-7}
Median = (-5 + -6)/2 = -11/2 (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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08 Jul 2026, 09:38

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The important information: 'n' is an integer. and the equation would be satisfied only when 'n' would be a negative integer.
As the middle term of the given inequality has to be greater than 30 and (1/3) would be inversed only from a power with a negative number.

So if we would check, 3^7 = 2187 (we should learn this number as it is frequently used, also could be derived from 9^2 * 3)
So 'n' max integral value is -7

and for n min. value the middle term shall have the value of n that just gives us greater number than 30.
3^3 = 27 and 3^4 = 81
So, 'n' min integral value is -4.

So 'n' has values [-7,-4]
These brackets depict that 'n' covers all values from -7 to -4

values of 'n' = -7, -6, -5, -4

The median would be -6-5/2 = -11/2

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08 Jul 2026, 09:38

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The minimum possible value of n is -7 and max possible value is -4. Hence the median is -5.5 which is -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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Since 3^-n is positive, it implies that n is negative integer.

List out all possible values 3^-n where it will be b/w 30 & 3000: {81, 243, 729, 2187 }
Corresponding values of n are: {-4, -5, -6, -7}

The median of this is: (-6-5)/2 = -11/2 Option D

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The inequality we've been given is : 3000 > 1/3^n >30
1/3^n can be written as 3^-n
Therefore, the above inequality can be expressed as 3000 > 3^-n > 30
Now, the powers of 3 that fall between 30 and 3000 are as follows:
3^4 = 81
3^5 = 243
3^6 = 729
3^7 = 2187
Thus, n can have 4 possible values i.e., -4, -5, -6 and -7.
The median of these 4 values is 1/2*[(-5) + (-6)] = -11/2.
Final answer: D. -11/2

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08 Jul 2026, 10:03

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The expression can be written as

30 < 1/3^n < 3000
i.e
30 < 3^(-n) < 3000

Only the following values will satisfy

3^4 = 81
3^5 = 243
3^6 = 729
3^7= 2187

Note the values of n will be negetive but since there is a negetive to cancel out the negetive sign outside n

Values beyond these would exceed the inequality

Hence median of -7, -6, -5 , -4 would be (-6-5) /2 which is -11/2 i.e Answer 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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3000 > 1/3^n > 30
3000 > 3^-n > 30; which implies 3^-n can be minimum 30 and maximum 3000.
n is also an integer which implies 3^-n is also an integer and a power of 3 between 30 and 3000

The possible values between 30 and 3000 which is a power of 3 are :
1. 81 = 3^4
2. 243 = 3^5
3. 729 = 3^6
4. 2187 = 3^7

Therefore, the values of 3^-n that works are n=-4,-5,-6,-7

Therefore, the median of all possible values of n is the average of -5 and -6 (even no of terms) = (-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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The question says 30< 1/3^n <3000

If n is +ve, there is no valid solution to the above question (because 1/3 can never fall between 30 and 3000 and if 1 is divided by an increasing +ve power, the answer becomes all the more impossible)

Hence n has to be negative which will simplify the question as 30<3^n<3000
Now the least power of 3 that's more than 30 is 3^4 (81) and the greatest power of 3 that is less than 3000 is 3^7 (2187). Hence the values of n can be -4,-5,-6,-7 and the median will be -11/2

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08 Jul 2026, 10:05

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For 1/3^n to be between 3000 and 30 n has to be negative so 1/3 becomes an integer.
N could be between (-4, -5, -6, -7) ------->
1/3^(-4)= 81
1/3^(-5)=243
1/3^(-6)=729
1/3^(-7)=2,187
Anything smaller than -7 will be greater than 3000
So median of (-7, -6, -5, -4) = (-6) +(-5)=-11 and -11/2
the answer is D
IMO

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given:
3000>(1/3)^n>30

(1/3)^n=3^-n

therfore: 30<3^-n<3000

powers of 3 = 3^7=2187<3000
3^8=6561>3000

so, 3^-n<3000= -n<=7

-n=4,5,6,7
n=-4,-5,-6,-7

ordered values -7,-6,-5,-4

-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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08 Jul 2026, 10:08

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3000>1/3^n = 3^-n>30
for 30:
n = -4(as 3^3 = 27 so we will take the next one)
for 3000 side:
n = -7(as 3^8 = 6561 which is greater that 3000, 3^7 will be approx 2k which is less than 3000)
so n = -4, -5, -6, -7
median = -5.5 = -11/2

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REALISE: that n has to be negatvie otherwise it wont be between the ranges. I just went through the numbers quickly, realised that cant be -3 as that would give 27, so start was 4 and then approximated the rest. 4,5,6,7, average is -11.5

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08 Jul 2026, 10:13

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Given:
1. n is an integer
2. 30 < 1/3^n < 3000

From second, we can infer that n has to be a negative integer (if it is 0 or greater than 0, 1/3^n will always be less than or equal to 1, hence not meeting the inequality)

So, based on the inequality, n's value is between -7 and -4 (3^4 to 3^7 is between 30 and 3000)

Since n is an integer, possible values are: -7, -6, -5, -4

Median of this set is = -11/2

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n has to be a negative integer to flip the number. 1/3^1 = 1/3, but 1/3^-1 = 3. 3^3 = 27 which is less than 30, as n has to be greater than 30, so the lower value of n is 4. and as 3^8 is greater than 3000 and we want below 3000, n has to be 7. So the values of n will be -4, -5, -6, -7. Thus median = (-5+(-6))/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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08 Jul 2026, 10:40

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for 1/3^n to be greater than 30, n must be negative.

then least n for 3^n >30 is 3^4 = 81
greatest n for 3000 > 3^n is 3^7 = 2187

so final integers which satisfy the given condition = -4, -5, -6, -7
median of these = (-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

This question was provided by GMAT Club
for the GMAT World Cup Competition

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