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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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Bunuel

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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

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Bunuel

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GMAT Club Official Solution:

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

We need:

3000 > 1/3^n > 30

Since 1/3^n = 3^(-n), we have:

3000 > 3^(-n) > 30

Divide the whole inequality by 3:

1000 > 3^(-n - 1) > 10

Now compare powers of 3 with 10 and 1000:

3^2 = 9, which is less than 10.

3^3 = 27, which works.

3^4 = 81, which works.

3^5 = 243, which works.

3^6 = 729, which works.

3^7 for sure will be more than 1000.

So:

-n - 1 = 3, 4, 5, or 6

Therefore:

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

The median is the average of the two middle values:

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

Answer: D.

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08 Jul 2026, 08:09

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3000> 1/3^n > 1/30
Take Reciprocal:
1/3000 < 3^n < 1/30
Write as powers of 3
3^-8 < 3^n < 3^3

So:
-8<n<3
Possible values:
-7,-6,-5,-4
Median
(-6+-5 )/ 2
=-11/2
Answer D

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i am going with option d
1/3^2 = 3^-2, 30<3^-2<3000
-4,-5,-6 and -7 satisfies and their median is -6+(-5)/2 = -11/2
option a incorrect - minimum value and not median
option b incorrect - it can be an answer if 8 is listed as a possibility of satisfied number in the set.
option c incorrect - single middle value not median
option e incorrect - similarly middle value not median

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3000> (1/3)^n > 30

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

UNDERSTAND THE BOUNDS FIRST.

30 lies between 3^3 and 3^4
Similarly 3000 lies between 3^7 and 3^8

but 3^3= 27, 3^4= 81, 3^5= 243, 3^6=729 and 3^7=2187
3^7<3^-n <3^4

but 30<3^-n<3000

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

Median= (-5+ (-6))/2 = -11/2

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$3000>(1/3)^n>30\\
so when n=-4 (1/3)^-4 =81>30\\
\\
consider 9^3=3^6=729...this can be multiplies only one more time for (2/3)^n<3000\\
\\
hence n ={-4,-5,-6,-7}\\
\\
median=-11/2$

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08 Jul 2026, 08:19

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3000>1/3^n>30
1/3^n = 3^-n
3000>3^-n>30

3^3=27
3^4=81
3^5=243
3^6=723
3^7=2187
3^8=6561

values are between 30 and 30000

-n=-4,-5,-6,-7
Meduan values=-11/2

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Here's my solution for this question

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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

given that n is an integer and condition

3000>(1/3)^n>30

this will be true for values of n -4,-5,-6,-7

median value of n (-5-6)/2 = -11/2

OPTION D ; -11/2 is correct

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Ans is Option D = -11/2

Since n is an integer, we can check by putting values of n. Positive values will not suffice since it will make the number smaller.

Trying negative n values, we get the reciprocal and only -4, -5, -6 and -7 satisfy the inequality, because if n=-3, it gives 1/3^(-3) = 27 which is not greater than 30 and n=-8 is 6561 which is bigger than 3000.

Median of all possible values of n which are -4, -5, -6 and -7 comes out to be -11/2. [Sum of middle terms (-5 + (-6))/2 or -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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I think the answer is D.

30 < 3^-n < 3000.

3^3 = 27 so min value is 4
3^4 = 81^2 = 3^8 = approx 6400 which is way more so lowering to 7 which gives 81 x 27 = 80 x 30 = 2400 approx which is valid

so possible values of n = -4, -5, -6, -7.

Median = -6-5/2 = -11/2

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08 Jul 2026, 08:22

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Given the inequality:
3000 > 1 / 3[sup]n[/sup] > 30

Using the negative exponent rule, rewrite 1 / 3[sup]n[/sup] as 3[sup]-n[/sup]:
3000 > 3[sup]-n[/sup] > 30

Since n is an integer, -n must also be an integer. We evaluate the integer powers of 3 to find which values fall within the range of 30 to 3000:

3[sup]3[/sup] = 27 (Too small)
3[sup]4[/sup] = 81 (Valid)
3[sup]5[/sup] = 243 (Valid)
3[sup]6[/sup] = 729 (Valid)
3[sup]7[/sup] = 2187 (Valid)
3[sup]8[/sup] = 6561 (Too large)

The possible integer values for -n are 4, 5, 6, and 7. Therefore, the possible values for n are:
n ∈ {-4, -5, -6, -7}

Arranging the set in ascending order yields {-7, -6, -5, -4}. The median of a set with an even number of values is the arithmetic mean of the two middle terms (-6 and -5):
Median = (-6 + -5) / 2
Median = -11/2

(D)

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

$1/3^8=3^{-8} = 1/6561 < 1/3000 < 3^n < 1/30 < 1/3^3 = 3^{-3} = 1/27$

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

The median of all possible values of n = (-6-5)/2 = -11/2

IMO D

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08 Jul 2026, 08:24

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With the given conditions, n starts at 4 since n^3 will be 27 and will end at 7 since 3^7 is 2187.

Thus median is (2nd+3rd)/2
= (5+6)/2 = 11/2

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We can safely do a trail and error approach since the answer set will be small,

consider 1/3 ^ n = x

Take a positive number, let's say, n = 1, we get x = 1/3, does not satisfy,
take n = -1, x = 3, we need more
take n = -3, x = -27, more
at n = -4, we have x = 81
at n = -5, we have x = 271,
at n = -6, we have x = 729
at n - -7, we have x = 2187 (we can that the next number falls outside our range)

we can stop here.

The required median is hence of the solution set: {-4, -5, -6, -7} is = (-5 + (-6)) / 2 = -11/2

IMO, 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
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08 Jul 2026, 08:27

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Correct answer: E. -5

Explaination:
since 3^3 = 27 and 3^7 = 2160
3000 > 1/3^n > 30 means (aproximately): 3^(n+7) > 3^0 > 3^(n+3)
in other words: n + 7 > 0 > n +3
this leaves us -4, -5 and -6 as possible values of n. Median is -5. The answer is E.

And we get to use these numbers aproximately because n is an integer. 3000 = 3^7,2 or 3^ 7,3 or something like around that, does not really matter.

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08 Jul 2026, 08:27

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1/3^n = 3^-n for reference lets take it as "a"

with that a > 30
3^3 is 27, so -n should start with 4

a<3000, lets look it this way 3000 = 3*10*10*10 which is greater than 3*9*9*9, the closest to 3000, so 3^7
-n final value will be 7

so n values ranges from -7.-6.-5.-4 - median = (-6-5)/2 = -11/2

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08 Jul 2026, 08:28

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3000>[1][/3^n] >30 , this means n is negative value ,
thus n belongs to [-7,-3],
therefor median will be -5.

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08 Jul 2026, 08:29

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the qn is basically asking as: 30 < 3^{-n} < 3000

3^3 = 27 (small)
3^4 = 81 (valid)
3^5 = 243 (valid)
3^6 = 729 (v)
3^7 = close to 2200 (v)
3^8 = close to 6600 (too high)

The valid exponent values for -n are 4, 5, 6, and 7.
therefore median of n [-4,-5,-6,-7]= -11/2

Ans D

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08 Jul 2026, 08:31

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(1/3)^n to be 30, n has to be -ve. as the options we can see.
3^3=27, so |n| > 3 and lets try |n|=6, 3^3 x 3^3 = 27 x 27 = 729 now if we again increment n then |n|= 7 and n^7= 2187
After this if we increment n it will pass over 3000. so |n|<8 i.e. 8>|n|>3
Since n is -ve, -8<n<-3;
the values for n are -7, -6, -5, -4
And the median would be (-6-5)/2 = -11/2
Option D should be the answer

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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⚠️ Important: GMAT Club does not allow AI-generated posts. AI-generated solutions are not eligible for kudos, and users who post them may face moderation action, including a ban.