Last visit was: 18 Nov 2025, 22:26 It is currently 18 Nov 2025, 22:26
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
 1   2   3   4   5   
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,104
 [21]
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,104
 [21]
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:00
1
Kudos
Add Kudos
20
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

52% (02:08) correct 48% (02:41) wrong based on 382 sessions

History

Date
Time
Result
Not Attempted Yet
If x is a positive integer and x^3 is a multiple of by 81, which of the following could be the remainder when x is divided by 36?

I. 0
II. 3
III. 27

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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

 

ShowHide Answer
Official Answer
D
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,104
 [3]
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,104
 [3]
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:30
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
If x is a positive integer and x^3 is a multiple of by 81, which of the following could be the remainder when x is divided by 36?

I. 0
II. 3
III. 27

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III
Show more

­

GMAT Club Official Explanation:



We are told that x is a positive integer and x^3 is a multiple of 81 = 3^4. If x^3 is divisible by 3^4, then x must be divisible by at least 3^2. In this case, x^3 = (3^2)^3 = 3^6, which is indeed divisible by 3^4. So, the least value of x is 3^2 = 9, and any multiple of 9 would make x^3 divisible by 3^4. So, for example, for each of 3^2, 3^2 * 2, 3^2 * 3, ..., x^3 will be divisible by 3^4.

Now we're asked which of the following could be the remainder when x is divided by 36. Since x is a multiple of 9, its remainder when divided by 36 could be:

  • 9, in case x = 9
  • 18, in case x = 18
  • 27, in case x = 27
  • 0, in case x = 36
  • 9, again, in case x = 45
  • and so on.

So, from the options, only 0 and 27 could be remainders.

Answer: D
General Discussion
User avatar
simondahlfors
User avatar
Joined: 24 Jun 2025
Last visit: 23 Sep 2025
Posts: 48
Own Kudos:
46
 [1]
Posts: 48
Kudos: 46
 [1]
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:16
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Information given:
- x is a positive integer
- x^3 is a multiple of 81
- So, x^3 must be divisible by 81 = 3^4

Question:
- Which of the following (1: 0, 2: 3, 3: 27) could be the remainder when x is divided by 36?

Solution:
- x^3 must have at least four 3's in its prime factorization
- So x must have at least 3^4/3, which means x must be at least x^2
- Therefore, x must be a multiple of 3^2 = 9

- Option 1: 0
- If remainder is 0 when divided by 36, then x is a multiple of 36
- 36 is a multiple of 9, so this satisfies the condition that x must be a multiple of 9, so possible
- Valid

- Option 2: 3
- If remainder is 3, then x = 36k + 3
- 36k is always divisible by 9, so 36k + 3 is never
- Invalid

- Option 3: 27
- x = 36k + 27
- 36k is divisible by 9, 27 divisible by 9, 27/3 = 3, so it's possible
- Valid

Answer: D, Option 1 & 3

Bunuel
If x is a positive integer and x^3 is a multiple of by 81, which of the following could be the remainder when x is divided by 36?

I. 0
II. 3
III. 27

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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

 

Show more
User avatar
MaxFabianKirchner
User avatar
Joined: 02 Jun 2025
Last visit: 12 Jul 2025
Posts: 65
Own Kudos:
62
 [1]
Given Kudos: 122
Status:26' Applicant
Location: Denmark
Concentration: Finance, International Business
Schools: ESSEC MiF '26 ESADE HEC Imperial LBS Oxford MFE '26 Sloan (MIT) NYU Stern NUS Bocconi IE'26
GPA: 4.0
WE:Other (Student)
Schools: ESSEC MiF '26 ESADE HEC Imperial LBS Oxford MFE '26 Sloan (MIT) NYU Stern NUS Bocconi IE'26
Posts: 65
Kudos: 62
 [1]
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:25
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
The problem states that `x` is a positive integer and `x^3` is a multiple of 81. The question asks for the possible remainders when `x` is divided by 36.

First, we analyze the condition that `x^3` is a multiple of 81. The prime factorization of 81 is `3^4`. For `x^3` to be a multiple of `3^4`, the prime factorization of `x^3` must contain at least four factors of 3. Let the number of factors of 3 in the prime factorization of `x` be `a`. Then the number of factors of 3 in `x^3` is `3a`. The condition `3a >= 4` must be satisfied. Since `a` must be an integer, the smallest integer value for `a` that satisfies this inequality is 2. Therefore, `x` must have at least `3^2` as a factor, which means `x` must be a multiple of 9.

Next, we determine the possible remainders when `x`, a multiple of 9, is divided by 36. Let `x = 36q + r`, where `q` is the quotient and `r` is the remainder. Since `x` is a multiple of 9 and `36q` is also a multiple of 9, the remainder `r` must be a multiple of 9. The possible remainders when dividing by 36 are integers from 0 to 35. The multiples of 9 in this range are 0, 9, 18, and 27.

We now check the given options against this set of possible remainders.
Statement I proposes a remainder of 0. This is possible. For example, if `x = 36`, it is a multiple of 9, and the remainder when 36 is divided by 36 is 0.
Statement II proposes a remainder of 3. This is not possible, as 3 is not a multiple of 9.
Statement III proposes a remainder of 27. This is possible. For example, if `x = 27`, it is a multiple of 9, and the remainder when 27 is divided by 36 is 27.

Since remainders of 0 and 27 are possible, statements I and III could be true. The correct answer is (D).
User avatar
Dav2000
User avatar
Joined: 21 Sep 2023
Last visit: 14 Sep 2025
Posts: 75
Own Kudos:
44
Given Kudos: 69
Posts: 75
Kudos: 44
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:26
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
If x is a positive integer and x^3 is a multiple of by 81, which of the following could be the remainder when x is divided by 36?

I. 0
II. 3
III. 27

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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

 

Show more
Given x is +ve integer and x^3 = 81 * n where n>1

x^3. = 3^4 * n
For x to be integer, n has contain 3^2 as a factor.
Therefore x = 3*3 * a where a > 1

Remainder of x/36 can be reduced to a/4
Therefore Case 1 and 2 that is 1 and 3 are possible But Case 3 which is 27 is not possible

There should have been an option of I and II only. SO for now let's select Option E.
User avatar
Mohak01
User avatar
Joined: 05 Sep 2020
Last visit: 18 Nov 2025
Posts: 104
Own Kudos:
64
 [2]
Given Kudos: 70
Location: India
Schools: ISB '26 NUS '26 (A)
GMAT Focus 1: 695 Q83 V87 DI83
GPA: 8
Schools: ISB '26 NUS '26 (A)
GMAT Focus 1: 695 Q83 V87 DI83
Posts: 104
Kudos: 64
 [2]
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:26
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Given, x is a positive integer and x^3 is a multiple of by 81.

Now, for x^3 can only be multiple of 81 if x is multiple of 9.

Hence possible values of x are multiples of 9, i.e., 9,18,27,36,45.... and so on
Further when the above possible values are divided by 36, remainders are as follows

For,
9 --> divided by 36 ---> Rem=9
18 --> divided by 36 ---> Rem=18
27 --> divided by 36 ---> Rem=27
36 --> divided by 36 ---> Rem=0
45 --> divided by 36 ---> Rem=9
.
.
... and so on, the probable remainders will repeat with a cycle of 4. Therefore only 0 and 27 form the above options are possible answers.

Therefore Option D is correct
User avatar
Heix
User avatar
Joined: 21 Feb 2024
Last visit: 18 Nov 2025
Posts: 361
Own Kudos:
153
 [1]
Given Kudos: 63
Location: India
Concentration: Finance, Entrepreneurship
Schools: Ross MM "26 NUS Columbia '27
GMAT Focus 1: 485 Q76 V74 DI77
GPA: 3.4
WE:Accounting (Finance)
Products:
My Reviews
Schools: Ross MM "26 NUS Columbia '27
GMAT Focus 1: 485 Q76 V74 DI77
Posts: 361
Kudos: 153
 [1]
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:31
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post

Show Spoiler
Attachment:
GMAT-Club-Forum-6byp1p1u.jpeg
GMAT-Club-Forum-6byp1p1u.jpeg [ 170.28 KiB | Viewed 1835 times ]
User avatar
Archit3110
User avatar
Major Poster
Joined: 18 Aug 2017
Last visit: 18 Nov 2025
Posts: 8,423
Own Kudos:
4,979
 [1]
Given Kudos: 243
Status:You learn more from failure than from success.
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1: 545 Q79 V79 DI73
GMAT Focus 2: 645 Q83 V82 DI81
GPA: 4
WE:Marketing (Energy)
GMAT Focus 2: 645 Q83 V82 DI81
Posts: 8,423
Kudos: 4,979
 [1]
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:35
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
81 = 3^4
for x^3 is a multiple of 81 ,
target
which of the following could be the remainder when x is divided by 36

option 1 0
36 = 2^2 * 3^2
x = 36 then 2^6 * 3^6
yes we can have remainder as 0

option 2
3
multiple value of 81 = 3^4 and that of 36 is 2^2 * 3^2
x has to be multiple of 9
not possible

option 3
27
27 = 3^3
(3^3 )^3 will be multiple of 81 and remainder will be 27 when divided by 36

sufficient

IMO option D is correct option I & III



Bunuel
If x is a positive integer and x^3 is a multiple of by 81, which of the following could be the remainder when x is divided by 36?

I. 0
II. 3
III. 27

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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

 

Show more
User avatar
Missinga
User avatar
Joined: 20 Jan 2025
Last visit: 16 Nov 2025
Posts: 393
Own Kudos:
261
 [1]
Given Kudos: 29
Posts: 393
Kudos: 261
 [1]
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:38
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
If x is a positive integer and x^3 is a multiple of by 81, which of the following could be the remainder when x is divided by 36?

81= 3^4 ; since x^3 is a multiple of 81, the x^3 must be divisible by 3^4 . The value of x must be more than (3^4)^(1/3) . This means x must be atleast divisible by 3^2=9. Hence x is 0,9 or a multiple of 9.
What remainders do multiples of 9 leave if divided by 36
0/36=0 remainder
9/36=9 remainder
18/36 =18 remainder
27/36 =27 remainder

From given choices I. 0..... II. 3......III. 27. Only 1 and III fit (0,27)

D
avatar
ManifestDreamMBA
avatar
Joined: 17 Sep 2024
Last visit: 18 Nov 2025
Posts: 1,282
Own Kudos:
784
 [1]
Given Kudos: 236
Products:
GMAT Club Tests
Posts: 1,282
Kudos: 784
 [1]
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:39
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
x > 0
x^3 = 81k, where k is some integer
x^3 = 3^4 *k

Given x is an integer, it should have atleast 2 3s

x= 9p, where p = cuberoot(3k)

x/36 = 9p/36

If p = 1, x/36 = 9
If p = 2, x/36 = 18
If p = 3, x/36 = 27
If p = 4, x/36 = 0
And the cycle will repeat after this

Answer I and III only

Bunuel
If x is a positive integer and x^3 is a multiple of by 81, which of the following could be the remainder when x is divided by 36?

I. 0
II. 3
III. 27

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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

 

Show more
User avatar
praneethkarempudi
User avatar
Joined: 23 Jul 2018
Last visit: 31 Oct 2025
Posts: 37
Own Kudos:
18
Given Kudos: 9
Location: India
Schools: ISB '26 IIMA '26 IIM
GMAT 1: 690 Q50 V34
GPA: 85%
Products:
My Reviews
Schools: ISB '26 IIMA '26 IIM
GMAT 1: 690 Q50 V34
Posts: 37
Kudos: 18
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:39
Kudos
Add Kudos
Bookmarks
Bookmark this Post
x^3 = 81 * (some value) = 3^4 (some value)
as x^3 is a perfect number, x^3 needs to be atleast 3^6* (something)
Therefore x is 9* (something)

Possible values of x:
1) 9*27 --> 9*27/36 --> reminder is 3
2) 9*4 --> 36/36 --> reminder is 0
3) 9*4*27 --> 36*27/26 -->reminder 27

Hence all 3 are possible.
Bunuel
If x is a positive integer and x^3 is a multiple of by 81, which of the following could be the remainder when x is divided by 36?

I. 0
II. 3
III. 27

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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

 

Show more
User avatar
iamchinu97
User avatar
Joined: 14 Dec 2020
Last visit: 18 Nov 2025
Posts: 132
Own Kudos:
139
 [1]
Given Kudos: 34
Products:
My Reviews GMAT Club Tests
Posts: 132
Kudos: 139
 [1]
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:46
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
We know that x^3 is multiple of 81 so we can write
x^3 = 81k
81k = 36p + R
72k + 9k = 36p + R

Now we can write 36 p = 36*2k = 72k
so then 9k = R
Now if K =0 then R = 0
if K = 3 then R = 27

Hence I and III is could be R
Hence Ans D
User avatar
Emkicheru
User avatar
Joined: 12 Sep 2023
Last visit: 12 Sep 2025
Posts: 119
Own Kudos:
22
Given Kudos: 11
Location: Kenya
Schools: Anderson (UCLA) Sloan (MIT) Columbia '27
GMAT 1: 780 Q50 V48
GRE 1: Q167 V164
GPA: 3.7
Schools: Anderson (UCLA) Sloan (MIT) Columbia '27
GMAT 1: 780 Q50 V48
GRE 1: Q167 V164
Posts: 119
Kudos: 22
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:46
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
If x is a positive integer and x^3 is a multiple of by 81, which of the following could be the remainder when x is divided by 36?

I. 0
II. 3
III. 27

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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

 

Show more
Let x be 27 therefore x^3 is 27*27*27/36=27*27*3/4 hence 3 is the remainder
User avatar
bart08241192
User avatar
Joined: 03 Dec 2024
Last visit: 17 Nov 2025
Posts: 75
Own Kudos:
64
 [1]
Given Kudos: 13
Posts: 75
Kudos: 64
 [1]
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:47
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
X is an element of the set of positive integers.
X^3 is a multiple of 81.
In other words,
X^3 is a multiple of 3^4.
81 contains four instances of 3.
Therefore, X must contain at least two instances of 3 to satisfy the condition of being a multiple of 81.

(3^2)^3 is the minimal expression that can be evenly divided by 81.

From this, it is evident that
X is a multiple of 9.
Numbers such as 9, 18, 27, 36, etc., meet the criteria.

When divided by 36:
9/36 yields a remainder of 9,
18/36 yields a remainder of 18,
27/36 yields a remainder of 27,
36/36 yields a remainder of 0.

Thus, remainders of 0 and 27 are compliant with the conditions.
User avatar
Cana1766
User avatar
Joined: 26 May 2024
Last visit: 15 Nov 2025
Posts: 85
Own Kudos:
79
 [1]
Given Kudos: 11
Posts: 85
Kudos: 79
 [1]
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:48
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Here x^3 must be a multiple of 81.
So 9*9*9 must come minimum for x^3 for it to be a multiple of 81.
So its 81*9[which is a multiple of 81]
When x/36 the remainder is 9.

next how about I multiple 2 with each 9
so 18*18*18 which should also be a multiple of 81(because there is still 9*9 in it)
when x/36 the remainder is 18

next 3 with 9
so 27*27*27
when x/36,the remainder is 27.So III is the ans.

next 4 with 9
so 36*36*36
when x/36 the remainder is 0.So I is the ans.

next 5 with 9
45*45*45
when 45/36,the remainder is 9.

So the answer is I and III.

Hence D is the answer
Bunuel
If x is a positive integer and x^3 is a multiple of by 81, which of the following could be the remainder when x is divided by 36?

I. 0
II. 3
III. 27

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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

 

Show more
User avatar
Rishika2805
User avatar
Joined: 13 Dec 2023
Last visit: 18 Nov 2025
Posts: 31
Own Kudos:
24
 [1]
Given Kudos: 27
Posts: 31
Kudos: 24
 [1]
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:49
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Lucky me, today I revised properties of number and I got this question today

x^3 is a multiple of by 81 means x^3 is divided by 81 that means x^3 contains at least four 3s = 3^4 only then 81 is divisible by x^3

So we can say x can contain 3 that means x^3 contains only 3^3 which has one less 3 so it is not divisible by 81

we can concluded x has at least two 3*3 because then x^3 will be (3*3*3*3*3*3) which is six 3s greater than four 3s

Then x could be 9, 18, 27,36.........

Now let's divide each number by 36 to find the remainder
9/36= 9 remainder, 18/36= 18 remainder, 27/36= 27 Remainder, 36/36= 0 remainder so answer is D (only 1 and 3)


Bunuel
If x is a positive integer and x^3 is a multiple of by 81, which of the following could be the remainder when x is divided by 36?

I. 0
II. 3
III. 27

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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

 

Show more
User avatar
chasing725
User avatar
Joined: 22 Jun 2025
Last visit: 17 Aug 2025
Posts: 85
Own Kudos:
81
 [1]
Given Kudos: 5
Location: United States (OR)
Schools: Stanford
Schools: Stanford
Posts: 85
Kudos: 81
 [1]
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:53
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
If x is a positive integer and x^3 is a multiple of by 81, which of the following could be the remainder when x is divided by 36?

I. 0
II. 3
III. 27

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III


 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

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

 

Show more

x is a multiple of 9. Hence we can write x = 9 * p, p is some constant.

x = 9*p

Remainder(x/36)

If p = 4 - the remainder can be zero
If p = 3 - the remained can be 27

For no values of p, the remainder can be 3.

Option D
User avatar
AviNFC
User avatar
Joined: 31 May 2023
Last visit: 13 Nov 2025
Posts: 216
Own Kudos:
288
 [1]
Given Kudos: 5
Posts: 216
Kudos: 288
 [1]
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:54
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
x^3=3^4*k
so x must be multiple of 9. x=9z

1. if z=4, x=36..remainder 0.. YES
2. 9z=36k+3..k cannot be integer...NO
3. if z=3..x=27..remainder =27..YES

Ans D
User avatar
BeachStudy
User avatar
Joined: 30 Jun 2025
Last visit: 18 Aug 2025
Posts: 61
Own Kudos:
37
 [1]
Given Kudos: 4
Posts: 61
Kudos: 37
 [1]
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 08:59
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
X is Positive Integer and a multiple of 81 when x^3


Solving: When you divide x by 36 the remainder could be.... 0, 3, or 27


x^4 ( 3*3*3*3) = 81

So since x^3 is divisible by 81 then [x^3][/3^4]. x must have (3*3) in it then and is divisible by 9.
Now we start to plug and play with the options:

0: if x is divisible by 36. 36/9=4 with no remainders. so 0 will work.
3: x= 36(number) + 3. since 3 can not be divided by 9 this does not work
27: x = 27(number) +27. 27 is divisible by 9 so this will work.

Answer: 0 and 27 - D
User avatar
amansoni5
User avatar
Joined: 16 Nov 2021
Last visit: 18 Nov 2025
Posts: 45
Own Kudos:
26
 [1]
Given Kudos: 97
Products:
Forum Quiz
Posts: 45
Kudos: 26
 [1]
GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 09:08
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
We need x^3 divisible by 81 =3^4 ⇒ x must have at least two factors of 3 ⇒ x divisible by 9

Check each remainder mod 36:

0 mod 36 ⇒ divisible by 36 ⇒ divisible by 9 ⇒ Valid.

3 mod 36 ⇒ examples: 3, 39 ⇒ not divisible by 9 ⇒ Invalid.

27 mod 36 ⇒ examples: 27, 63 ⇒ divisible by 9 ⇒ Valid.

Answer: D. I and III only
 1   2   3   4   5   
Moderators:
Bunuel
Math Expert
105355 posts
Krunaal
Tuck School Moderator
805 posts