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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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Reon

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08 Jul 2026, 08:32

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3000>3^(-n)>30
3^3=27
3^4=81
3^5=243
3^6=729
3^7=2,187
3^8=6,561
The possible values for 3^(-n) are 81, 243, 729, 2187.
Possible values of n = -4, -5, -6, -7
Median = -5-6/2 =-11/2

D. -11/2

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[quote="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

Given n = integer and 3000> 1/3^n > 30

Now one this is clear from this inequality given that n has to be negative and also it is confirmed by options in answer choices.
Now let us find values of power of 3 one by one respecting this inequality above:-

3^3=27 ❌
3^4=81 ✅
3^5=243✅
3^6=729✅
3^7=2187✅
3^8=6,561 ❌
Only 4 powers of 3 satisfy this ie 4 values of n = 4,5,6,7
Now as no. Of terms are even therefore median = average
ie (5+6)/2 =11/2
And as n should be negative therefore median = -11/2

IMO correct option (D)

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3000 > 1/3^n > 30

n is an integer

n has to be negative integer so 3 can become a numerator to satisfy the equation. If n is positive, equation breaks. Now solving:

n = -1 value of 1/3^n = 3; 3 is not greater than 30, we need minimum value of n to be -4, that gives 81 - equation satisfied. lets go further - equation works at -5, -6, -7 at -8 the value is 6561, equation doesnt hold. we now have all values, arranging in ascending order -7, -6, -5, -4, median = -6-5/2 = -11/2 - Answer D

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08 Jul 2026, 08:36

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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

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Its given 30<1/3^n<3000
So when
1/3^-4 = 81
1/3^-5=243
1/3^-6=729
1/3^-7=2187
Any number value for n below 4 and 7 would not be valid according to the condition.
So the n={-4,-5,-6,-7}
Hence median would be [-5+(-6)]/2=-11/2
Answer : D

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08 Jul 2026, 08:38

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3000>1/3n>30
30<1/3n<3000
exponent rule 1/3n = 3^-n
30<3^-n<3000
powers of 3 to see which one is btwn30 to 3000
3^3=27 too small
3^4=81 fits
3^5=243 fits
3^6= 729 fits
3^7= 2187 fits
3^8=6561 too big

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

median
4 numbers median is average of 2 middles
-6+(-5)/2 = -11/2
ans:D

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08 Jul 2026, 08:38

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Answer (D) -11/2

3000>$1/3^n $> 30

$3^3= 27$
$3^4 = 81$ hence n starts at -4
$3^5 = 243$
3^6= 7xx
3^7 = 21xx hence n ends at -7
3^8 = 6xxx which is greater than 3000

Set = {-4,-5,-6,-7}
Median $= (-5-6)/2 = -11/2$

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08 Jul 2026, 08:39

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We are given, 3000 > 1/(3^n) > 30. This means, 1/3000< 3^n< 1/30

We know 3^3 is 27, so 1/27 would be greater than 1/30. Since n is an integer, next value would n=4, which 3^4 and 1/81 (3^-4) is less than 1/30. Similarly, we see 3^7 is 2187 (3^-7) so 1/2187 is greater than 1/3000. n>7 would fetch values less than 1/3000. Hence n lies between, -7 and -4. That is n lies between: -7,-6,-5,-4. Median is between -5 and -6 which is -11/2.

Hence answer is D.

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1/(3^n) = 3^-n , so re writing the equation : 3000 > 3^-n > 30
max values of n -> 3^4=81, 3^7=729
possible values of n = -4,-5,-6,-7
median = -5-6/2 = -11/2

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if 3^n is in the denominator, n would have to be negative for it to fit in the equation (negative power would bring the 3 in the numerator to become positive). now if we take 3^3 = 27, which is less than 30, n cannot be -3, n has to be greater than -3, so possible values of n would be -4, -5,...

to get the upper limit, do trial and error, 3^5 = 243, when multiplied by 3^3 = 27 (243*27) gives us 6561, hence we try to do 3^4 * 3^3 = 3^7 = 81*27 which gives us 2187 which is under 3000, hence fits the equation. hence possible values of n = -4, -5, -6, -7 (has to stop here to fulfil the equation)

median = average of middle 2 terms = (-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, 08:49

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Answer: D) -11/2

1/3^n can be rewritten as 3^-n

We know that we need a negative value of n so that it becomes a positive exponent of 3 and yields a value between 30 and 3000.

We know the exponents of 3 that yield values between 30 and 3000: 3^4, 3^5, 3^6, and 3^7.

The median of 4,5,6,7 is 5.5, which is 11/2. The negative value is -11/2.

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If 3000>1/3^n>30 then n must be negative because otherwise a fraction would not be bigger than 30.

So to evaluate this I changed the inequality to 3000 > 3^n > 30 and just remembered that all my n values need to be negative when I evaluate at the end.

All n values are integers so I listed different values of n that resulted in 3^n being greater than 30 but less than 3000

3^4 = 81
3^5 = 243
3^6 = 729
3^7 = 2187
3^8 = 6561 which is two big so I can stop there.

The numbers need to be negative so for the set of n I have -7, -6, -5, -4. Since their are an even number of terms I need to take the sum of the middle two terms and divide by 2 so ((-6) + (-5))/2 = -11/2, so D is the answer.

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08 Jul 2026, 08:55

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-4,-5, -6 these satisfy so the answer is -5

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MY WAY TO FIND: GIVEN THAT n is INTEGER

CHECK FOR n=0: 3^0=1 value b/w 3000 to 30: NO
CHECK FOR n=1: 3^1= 3 value b/w 3000 to 30: NO
IF WE FURTHER INCREASE POSITIVE INTEGER POWER THEN THE DENOMINATOR WILL INCREASE AND VALUE DECREASE.. SO IT CAN ONLY BE -VE INTEGER

CHECK FOR n=-1" 3^-1= 1/3 but again 3 donot satisfy the condition

similarly when n=-4 then 3^-4=1/81 then only it satisfies the condition and it satisfy till n=-7
again at n=-8 it goes beyond so it does not satisfy

so integer values that satisfy conditions are -4,-5,-6,-7 and median is (-5)/2+(-6)/2=-11/2

Answer-D

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08 Jul 2026, 09:04

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The answer is D.

First, let's rewrite the given inequality to make it easier to read:

30 < 3^-n < 3,000

We need to find all values of n for which the above inequality is true. Note we must take negative values of n because the negative 1 before the exponent will turn it positive.

The greatest value of n will be -4, since 3^4 = 81 > 30.

To find the least value of n, I simply kept increasing by powers of 3...
3^5 = 243 (memorized)
3^6 = 729
From here, we can estimate.
3^7 ~ 2200
And we see that 3^8 will certainely be too large.

Therefore, the list of values is -4, -5, -6, and -7.

The median of this list is the average of the two central values, i.e., -5 and -6. Therefore the answer is -11/2, D.

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Given ,
30< 1/3^n < 3000
or, 30< 3^-n < 3000

If , n = 1 , 3^-1 =1/3 , not valid as 3 ^-n must be bigger than 30 .

so , the value of n is negative.

if n= -1 , 3 ^ -(-1) = 3 , not valid
if n=-2 , 3^-(-2) =9, invalid
if n = -3 , 3^-(-3) =27 also invalid
if n=-4 , 3^-(-4)=81 valid as 3^_n must be greater than 30
if n=-5 , 3 ^-(-5)=243 valid
if n =-6 , 3^-(6)= 729 valid
if n=-7, 3 ^-(7)= 2187 valid
if n =-8 , 3^-(-8)= 6561 invalid as 3^n must be less than 3000

So , -4 and -7 are the lowest and highest threesold .
Hence a set of possible values of n can be N ={ -4 , -5 ,-6,-7 }

as the number of the elements of N is even , the median is ,

-5 + (-6) /2
= -11 /2
So ans is option d . -11/2

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08 Jul 2026, 09:07

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What is given to us:
n is an integer
30< 1/3^n < 3000

Now the first deduction looking at the answer choices and the information given is that all values for n have to be negative. If n is positive 1/3^n will always be less than 1 and will never satisfy the fiven condition. When n is negative (for eg: -2) 1/3^(-2) = 3^2

Now, listing powers of 3
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243
3^6 = 729
3^7 = Some value in the range of 2000-3000
3^8 = Some value above 6000.

Thus we know, the set of values of n = {-4,-5,-6,-7)

Median will be the average of -5 and -6 which is -11/2.

Hence answer is (D)

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08 Jul 2026, 09:08

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x can take -4,-5,-6 median -5

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The answer is D here. 3^4 = 81, 3^5= 243, 3^6 = 729 3^7 = 2187 . So median would be -5 and -6 here as and it would be -11/2 .

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08 Jul 2026, 09:21

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My answer is D.) -11/2
3000 > 3^-n > 30

3000 can be written in the form of 3^n
3*10^3> 3*9^3
3*(3^2)^3= 3^7
3000>3^7 and because of less than sign we can say that 3^7 is part of the range hence 3000 ~ 3^8

30>3^3 and because of greater than sign we can say that 3^3 is not part of the range hence 30 ~ 3^3

3^3 < 3^-n < 3^8. Therefore, n can be -4,-5,-6,-7. 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

This question was provided by GMAT Club
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To solve:
3000 > 1/3^n > 30

So,
3000 > 3^-n > 30

Comparing the power of 3, we have,

3^3 = 27 (Not ok)
3^4 = 81 (ok)
3^7 = 2187 (ok)
3^8 = 6561 (Not ok)

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

Median of these values are,
= (-6-5)/2 = -11/2

Answer: 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.