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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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08 Jul 2026, 10:46

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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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To satisfy the inequality, n can take the following values: -4, -5, -6, -7. Hence, the median is the middle of 2nd and 3rd terms, which is -11/2.

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08 Jul 2026, 11:14

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i think option D is the option !!
--- 3000 < 3 ^ - n < 30
here n can't be less than more than -4 (-3 gives us 3^3 Which is less than 30 )and can't be less than -7 ( same -8 gives us 3 ^ 8 Which is more than 3000 )
--- So the possible values are - 4 , - 5 ,- 6 and - 7 so here the median is so its -22 / 4 so its -11 /4
----- That is my approach if anyone has any way to solve feel free to share : )

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08 Jul 2026, 11:15

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Rewrite the original expression as:

30 < 3^(-n) < 3000

Now compare with powers of 3.

Since 3^3 = 27 ad 3^4 = 81, -n must be at least 4.

For the upper limit, use approximation

3^7 = 3^4 * 3^3 = 81*27 ~ 80*30 = 2400
3^8 = 3^4 * 3^4 = 81*81 ~ 80*80 = 6400

Since 3000 lies between these 2, -n must be less that 8.

Possible values of n are: -7, -6, -5, -4

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

Ans : D

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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

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

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given,
$3000>3^-n>30$

lets find out possible values of n
if n = -3, $3^-(-3)=27$ but 27<30 so not valid
n=-4, $3^-(4) = 81 $ 3000>81>30 valid
n=-5, $3^-(-5) = 243 $ valid
n=-6, $3^-(-6) = 729 $ valid
n=-7, $3^-(-7) = 2187$ valid
n=-8, $3^-(-8) = 6561 $ but 6561>3000 - so invalid
possible values of n : -4, -5, -6, -7
median : $\frac {-5-6} {2} = \frac {-11} {2} $
-11/2
D

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08 Jul 2026, 11:36

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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

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

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We can say that

3000 > 3^(-n) > 30
It is clear that n would be in a negative range
3^7 = 2189 and any higher power would make the result greater than 3000
similarly 3^3 = 27. Since the range is > 30 at the lowest, we can consider the power of 4 as 81 > 30
Hence the resulting equation becomes:
7 > -n > 4
-7 < n < -4
The numbers in the range are -5 and -6
average of both is -5.5 = -11/2
Hence answer is D. -11/2

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08 Jul 2026, 11:46

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3000 > 1/3^n > 30

Rephrasing 30< 3^-n < 3000

Listing out powers of 3

3^1 = 3, 3^2 = 9, 3^3 = 27, 3^4 = 81 , 3^5 = 243, 3^6 = 729, 3^7 is around 2300, 3^8 is around 6900.

S0 3^-n needs to be between 3^4 and 3^7

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

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

Answer D

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08 Jul 2026, 11:49

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The question tells us:

3000>(1/3^n)>30
n is an integer

We need to find the median of all possible values of n.
Its important in such questions to take a moment and plan a course of action before attacking the question.

Let's rewrite the given inequality as:
3000> 3^(-n) >30

Since 3000 is a big number, and finding the power of 3 under 3000 seems like a daunting task, so let's make it a little easier by dividing all terms by three. We get:

1000 > 3^(-n-1) > 10
Now, finding the lower limit is easy, 3^3>10>3^2
Therefore, least possible value of (-n-1) = 3
-n=4
n=-4 (Least possible integer value of 'n')

For the highest possible value, let's do some estimation:
Since 1000 = 100*10, and we need value less than 1000
The closest values lesser than 1000, that are also powers of three are are 81 & 9.
81*9 gives us 729.
For safety, we quickly check that if we were to increase the power by one more, what will happen, and we can see that it will take the product well over 2100 (since 700*3=2100)

Therefore, highest value of 3^(-n-1)=
81*9
=(3^4)*(3^2)
=(3^6)

For highest possible value of 'n':
-n-1=6
-n=7
n=-7.

This gives us four integer values of 'n':
-4,-5,-6,-7

Since we get four possible values, median will be:
[-5+(-6)]/2
=-11/2

Answer: D

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08 Jul 2026, 12:02

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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

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

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3000>3^-n>30
3^3=27(low), 3^4=81, 3^5=243, 3^6=729, 3^7=2187, 3^8=6561(too high)
to fit in the given range...n should be in the range of 4 to 7
3^3<3^-n<3^8
3<-n<8
-3>n>-8
which means possible values of can be : -4, -5, -6, -7
therefore, median of these will be -11/2

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Refer attached working

Attachments

photo_2026-07-09_00-33-19.jpg [ 161.97 KiB | Viewed 98 times ]

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08 Jul 2026, 12:29

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y= 1/3^n (for simplicity of writing answer)

n = -3, y= 27
n=-4, y=81
n=-5, y=243
n=-6, y=700ish
n=-7, y=2100ish

we know now that n=-8 will exceed 3000, so we have our possible values.
{-4, -5, -6, -7}

The Median will be the avg of 2nd and 3rd term --> [(-5) + (-6)] / 2 --> -11/2

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08 Jul 2026, 12:34

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3000 > 1/3^n > 30

3000 > 3^-n > 30

so 30 < 3^-n < 3000

3^3 = 27
3^4 =81
3^5=243
3^6=729
3^7=2187
3^8=6561

therefore 4 less than -n which is less than 7

therefore n is -7,-6,-5,-4

median is -5 and -6 divided by 2 = -11/2

Answer D -11/2

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08 Jul 2026, 12:37

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Start with -4 as integer and not 4 as it is inverted, reciprocal, now 1/3^-4 is actually 81, then -5 would be 243 and if you keep going maximum will be -7, now the median here is between -5 and -6 which will be -5-6/2 which is -11/2. Which is option D.

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08 Jul 2026, 12:52

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3000>3^(-n)>30

30 > 27
and 27 = 3^3
Also, 3^7 = 2187 and 3^8 =6561
Thus, possible values for -n are {4,5,6,7}
or, posisble values of n are {-7,-6,-5,-4}
Median = {(-6)+(-5)}/2 = -11/2
Hence, correct ans: D

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08 Jul 2026, 13:18

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if 1/3^n has to be more than 30 and less than 3000
Then n has be negative power

Now,
3^4 = 81 (more than 30)
and 3^7= 2187 (less than 3000)
so n will have only 4 values that are -4, -5, -6, -7
Now median of these will be
{ -5 + (-6) }/2 = -11/2
Hence, D

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08 Jul 2026, 13:35

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Answer should be D = -11/2

3000> 1/3^N > 30

Since 1/3^n = 3^-n, we can write the inequality as --> 30 < 3^-n < 3000

Looking at powers of 3

3^3 = 27
3^4 = 81
3^5 = 243
3^6 = 729
3^7 = 2187
3^8 = 6561

Only numbers that qualify as n are 4, 5, 6, and 7.

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

Median (avg of middle 2 numbers in case of even count) = (-6-5)/2 = -11/2 (option D)

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08 Jul 2026, 14:05

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We can express it as 3000 > 3^(-n) > 30.
3^3 is 27 and 3^4 is 81 so min number is 4.
3^7 is 2187 and 3^8 is 6561
So min is 4 and max is 7 for values of (-n).
Median value of n = (-7 - 4)/2 = -11/2
Ans is D IMO

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

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

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08 Jul 2026, 14:06

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powers of 3: 3,9,27,81,243,729,2187

3000>1/3^n ==> 3000>3^(-n) == min interger value of n >=-7 as 3^7=2187
similarly 1/3^n > 30 ==> 3^(-n)>30 ==> n<=4 (as 3^4= 81 & 3^3 =27)
median n = (-4-7)/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

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

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08 Jul 2026, 14:32

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For 1/3^n to be greater than 30, n>-3.
n=-4. 81
n=-5, 243
n=-6 729
n=-7 2187, so n=-8 cannot possibly be less than 3000.

Median = (-5-6)/2 =-11/2 Ans D.

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08 Jul 2026, 15:02

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3^3 = 27 so it doesn't fit the >30 range, so we must start at 3^4 = 81, keep multiplying by 3 and you will see that the limit is 3^7
so n can be -4, -5, -6, -7 and median is then -5.5

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I found the best solution to plug and chug through this question. Looking at the equation in the middle, I knew the values for n needed to be negative ( I could also look at the answer choices...). Plugging in -1, I can solve for 3. This is too small so I kept going.

-2---> 9
-3 --> 27
-4--->81
-5-->243
-6 -->729
-7-->2,187

Since n is an integer, I can take the average of the middle two terms which gives me answer choice D, -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

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

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

⚠️ 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.