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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 09:21
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\(x^3 =\) multiple of 81
=> \(x^3\) must be divisible by \(3*3*3*3 = 3^4\)
=> \(x\) must be a multiple of \(9 (3*3)\) to have at least four 3's in \(x^3\)
=> \(x = 0, 9, 18, 27, 36,....\)
=> when \(x\) is divided by 36, the remainders can only be multiples of 9

Therefore, out of the three options given in the question, only I and III are possible. Hence, the answer is D.
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


 


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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 09:35
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If x is a positive integer and x^3 is a multiple of 81, which of the following could be the remainder when x is divided by 36?

x^3 is a multiple of 81 = 3^4; 3^4 is a factor of x^3; 3^2=9 is a factor of x.

x = 9k

k = 0; The remainder when x=0 is divided by 36 = 0
k = 1; The remainder when x=9 is divided by 36 = 9
k = 2; The remainder when x=18 is divided by 36 =18
k = 3; The remainder when x=27 is divided by 36 = 27
k = 4; The remainder when x=36 is divided by 36 = 0
...
The remainders when x is divided by 36 = {0,9,18,27}

I. 0; Possible
II. 3; Not possible
III. 27; Possible

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

IMO D
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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 09:40
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We are told that x^3 is divisible by 81 = 3^4.

This means,
x must be divisible by at least 3^2 = 9 ,
because (3^2)^3 = 3^6
which is enough to ensure that x^3 is divisible by 81.
So, x must be divisible by 9.

Now, check each option to see if x could be divisible by 9:

Option I: Remainder 0 when divided by 36
Then x=36k, which is divisible by 9.
Valid.

Option II: Remainder 3 when divided by 36
Then x=36k+3,
which is not divisible by 9.
Invalid.

Option III: Remainder 27 when divided by 36
Then x=36k+27,
which is divisible by 9.
Valid.

Correct answer: D. I and III only
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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 09:41
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Given condition,
x is a positive integer and x^3 is a multiple of 81.
First, let's find the prime factorization of 81: 81=9×9 =3 ^4
Since x^3 is a multiple of 81, we can write x^3 =81k for some integer k.
x ^3 = 3^4* k

For x^3 to be a multiple of 3^4, the prime factorization of x^3 must contain at least 3^4
Let the prime factorization of x be x=3^a*m, where m is an integer not divisible by 3.
Then x ^3 =(3^3a*m^3)

For x^3 to be a multiple of 3^4 , we must have 3a greator than or equal 4.
Since a must be an integer, the smallest possible value for 3a that is greater than or equal to 4 is a=2, because 3×2=6.
If a=1, 3a=3, which is not greator than or equal 4.
So, a must be at least 2.

This means that x must be a multiple of 3^2 =9.
So, x=9n for some positive integer n.

Now we need to find the possible remainders when x is divided by 36.
Since x=9n, we can write x as:
If n=1, x=9. Remainder when 9 is divided by 36 is 9.
If n=2, x=18. Remainder when 18 is divided by 36 is 18.
If n=3, x=27. Remainder when 27 is divided by 36 is 27.
If n=4, x=36. Remainder when 36 is divided by 36 is 0.
If n=5, x=45. Remainder when 45 is divided by 36 is 9.
If n=6, x=54. Remainder when 54 is divided by 36 is 18.
If n=7, x=63. Remainder when 63 is divided by 36 is 27.
If n=8, x=72. Remainder when 72 is divided by 36 is 0.

The possible remainders when x is divided by 36 are 0, 9, 18, and 27.

Let's check the given options:
I. 0: Yes, this is a possible remainder (e.g., when x=36).
II. 3: No, this is not a possible remainder.
III. 27: Yes, this is a possible remainder (e.g., when x=27).

Therefore, the possible remainders are I and III
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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 09:45
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x^3 is multiple of 81 => x^3 is divisible by 81 => x is divisible by atleast 3^(4/3) or atleast 3^2 ie 9

we need to find x=0 mod 36 with condition x=0 mod 9
therefore all multiples of 9 between [0,35] will be the reminder for the required question, which are 0, 9, 18, 27

Hence I and III could be reminder
Answer D
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


 


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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 09:59
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If x^3 is a multiple of 81, then x^3 = n*81, where n is an integer.

81 = 9 * 9 = 3^2 * 3^2 = 3^4.

Since 81 is 3^4, our x^3 must have at least "four 3's". If x = 3^1 then x^3 = 3^3, that is not enough. If x = 3^2 them x^3 = (3^2)^3 = 3^6 and now we have enough "3's" to have a multiple of 81.

So, x must also be divisible by 9 (3^2).

Therefore, x = p*9.

Possible values of x (and remainder when mod 36) = 9 (9), 18 (18), 27 (27), 36 (0), 45 (9), 54 (18), 63 (27), 72 (0), 81 (9), 90 (18),...

We can see that the pattern of remainders, for all multiples of 9 divided by 36, is: 0, 9, 18, and 27.

Because of that, the possible remainders are (I) 0 and (III) 27.

Answer:
D. I and III only
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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 10:00
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Since x^3 is divisible by 81, and 81 is made from several 3s multiplied together, x itself must be divisible by 9 because when you cube something, any factors it has get tripled. That means x must be a multiple of 9. Now, when you divide a multiple of 9 by 36, the remainder you get could be 0 (if 36 fits in exactly), or it could be 9, 18, or 27 depending on how far past a full multiple of 36 the number goes. So, out of the choices listed: 0, 3, and 27 the ones that are possible remainders when dividing by 36 are 0 and 27. The number 3 wouldn’t work because 3 is not a remainder you get when dividing a multiple of 9 by 36.

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


 


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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 10:02
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x^3 is divisible by 81=3^4 so x must be divisible by 9
- x is divided by 36
- must be multiple of 9
- possible reminder 0,9,18,27

I 0 (x=36
36^3=46656= 81*576) true
II 3 is not multiple of 9 not ture
III 27 (x= 27 27^3= 19683= 81*243 ture

ans is D( I & III only)
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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 10:04
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Given, x is a multiple of 81.
So, \(x^3 = 81a = 3^4 a\) ; where a\(\to\) Interger
\(\implies x = 3^{4/3}a\)

So, x must be divisible by at least 3^2.
Since \(x^3\) has 4 3s in its prime factorization, x must have at least 2 3s for it to be an integer.
So, x must be divisible by 9.

Now, we have to find which of the numbers can be the remainder of x/36, given that x is divisible by 9.
We can just check which of the options are divisible by 9.
0 and 27 are the only ones that work

Answer: D) I and III only
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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 10:10
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81 = 3^4
For x^3 to be divisible by 3^4, x must be divisible by 3^2 = 9
(Because (3^2)^3 = 3^6, which is ≥ 3^4, but (3^1)^3 = 3^3 < 3^4)

When x is divided by 36, the remainder must be less than 36.
Since x must be divisible by 9, the remainder when dividing by 36 must be a multiple of 9 less than 36:
0, 9, 18, or 27

Check given options
I. 0 → possible
II. 3 → not possible (3 is not a multiple of 9)
III. 27 → possible
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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 10:16
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\(x\) > 0 integer and \(x^3\) is a multiple of 81

We can write, \(x^3 = 81m\); where \(m>0\) integer

Note, that \(81 = 3^4\), for x to be an integer \(x^3\) must have powers of all its prime numbers in multiples of 3; thus, m is contributing at least a \(3^2\) tp ensure that x is an integer

So, we can infer that \(x = 3^2 (m^\frac{1}{3})\)

Let's call \(m^\frac{1}{3}\) as \(p\),

we can say that \(x = 9p\)

x can be 9, 18, 27, 36, etc.

When x is divided by 36, the following are the remainders that we get, 9, 18, 27, 0 (Note that beyond this point for any value of x, the remainders repeat this order)

Out of the remainders that we got, only 27 and 0, match the answer choices.

So Answer is D.
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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 10:17
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X^3 = 81*K = 3^4*K

then X is at least a multiple of 3^(4/3 ) = ~=3^2 = 9

1. R = 0, Means X is divisible by X, 36 = 9*4, cantain 9
X can be = 9*4*k, could be, true

2. R = 3, then x is 36*K+3
Let's take some exp. X = 3, 39, 75, 111, .... 363
in all the exp there is not a single number where 36*K+3 and is a multiple of X match, then it's not possible

3. R = 27, then x is 36*K+27
Exp. X = 27, 63, 99.... all are multiples of 9, then it's possible

Ans I and III
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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 10:27
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Ans: D (I and III only)

If x is a positive integer and x^3 is a multiple of 81 = that means x^3 is a multiple of 3^4
So x is a multiple of 3 and (3a)^1/3
For x to be a positive integer, we know that (3)^1/3 can not exist, and a has to be a multiple of 3^2 to make it 3^3 so that x^3 becomes a multiple of 81*3 = 3^6
From this, we know that x is a multiple of 9, or in other words, x is divisible by 9.

Now the question: which of the following could be the remainder when x is divided by 36?
Let's say that the remainder is r then we can write x as
x = 36n + r and this has to be divided by 9 completely

I. 0 = 36n + 0 = 36 (divisible by 9) Correct
II. 3 = 36n + 3 (not divisible by 9) Incorrect
III. 27 = 36n+ 27 = (divisible by 9) Correct

D. I and III only (answer)
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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 10:50
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X^3 is a multiple of 81

Thus X^3= 81*A= 9*9*A
A should be 9 or 9*Cubes of any positive integers

Minimum X^3=9*9*9
Divided by 36-->(9*9*9)/36 (remainder is 9)

next Possible value of X^3= 9*9*9*(2^3)=9*9*9*8
Divided by 36--->(remainder is zero)

next Possible value of X^3= 9*9*9*(3^3)=9*9*9*27
Divided by 36--->(remainder is 3)

Remainder cannot be 27 because denominator and numerator have 9 in common. Since 36 in 9*4, 9 in numerator and denominator will always cancel out leaving only 4 in denominator (divisor). So the maximum remainder possible is 4 or below.

Hence only options possible are 0 and 3.
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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 11:03
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If \(x^3 \)is a multiple of 81, then\( x^3 \)is divisible by \(3^4\). Since \(x^3\) has three factors of x, it follows that x must be divisible by \(3^(4/3)\)=\(3^2\)=> 9. In other words, x must be a multiple of 9.

Now we will check from options:
I. 0
Possible

II. 3
not Possible , since it’s not a multiple of 9

III. 27
Possible

Hence, answer is 1 and 3 only.

Option D.
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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 11:42
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It's given that X^3 = 81k

Therefore x=3 root 3 k

For 3 to be an integer the minimum value of k will be 3^2 therefore 9 is the minimum possible number. Given that we can take k to be anything, 9*4=36 is a possible option so remainders of 0 and 27 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


 


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x^3=81k=(3^4)k (where k is a positive integer),
So, x has to atleast multiple of 9
x=9m (where m is a positive integer)

Lets, take some cases,
x can be 9, 18, 27, 36, 45, ..

For x=9, the remainder when x is divided by 36 is 9
For x=18, the remainder when x is divided by 36 is 18
For x=27, the remainder when x is divided by 36 is 27
For x=36, the remainder when x is divided by 36 is 0
For x=45, the remainder when x is divided by 36 is again 9
...And, this will keep on repeating..

So, IMO answer D
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


 


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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 12:00
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The Key Condition: x is a Multiple of 9
Since x^3 is a multiple of 81, and 81=3^4, x^3 must have at least four factors of 3. For x^3 to have at least four factors of 3, x itself must have at least two factors of 3. So, x must be a multiple of 9.

I. Remainder 0: If x leaves a remainder of 0 when divided by 36, then x is a multiple of 36. Since 36=4×9, any multiple of 36 is also a multiple of 9. This is possible.

II. Remainder 3: If x leaves a remainder of 3 when divided by 36 (e.g., x=39), then x can be written as 36k+3. This means x is not a multiple of 9 (it would leave a remainder of 3 when divided by 9). This is not possible.

III. Remainder 27: If x leaves a remainder of 27 when divided by 36 (e.g., x=27 or x=63), then x can be written as 36k+27. Since 36k is a multiple of 9 and 27 is a multiple of 9, their sum x is also a multiple of 9. This is possible.

Regards,
Lucas
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


 


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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 12:04
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x^3 is a multiple of by 81

this means x^3 is divisible by 81

81 = 3^4 , so x must be multiple of 3^2 because then only when we take a cube of x we will get 3^6 in x^3

therefore we know x is a at least multiple of 9

checking divisibility of x by 36

x = 36q +r

if r = 0 ; x = 36q, it has 9 as a multiple so zero remainder is possible

if r = 3 ; x = 36q+3 , 9 cannot be taken common, not possible

if r = 27 ; x = 36q+27 , 9 can be taken as common , 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


 


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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is 07 Jul 2025, 12:07
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If \(x\) is a positive integer and \(x^3\) is a multiple of \(81\), then we can write \(x^3\) as
\(x^3=81k\) (where k is any positive integer)
\(x = (81k)^{\frac{1}{3}}\)
\(x = 3(3k)^{\frac{1}{3}}\)

We can now set \(x\) equal to those numbers which give the remainders we need when divided by \(36\).

(1.) \(0\) -> To get \(0\) as a remainder, \(x\) has to be a multiple of \(36\).

\(x = 36a\) (where \(a\) is any positive integer)

\(3(3k)^{\frac{1}{3}}=36a\)

Cube both sides

\(3^4k = 36^3a^3\)

\(k = \frac{36^3a^3}{3^4}\)

\(k\) is clearly an integer because \(81 = 3^4\) and \(36^3\) has enough \(3\)'s in it's prime factorization to cancel the denominator.

\(k\) came out to be an integer, that means there is a value of \(x\) which is completely divisible by \(36\).

(2). \(3\) -> To get a remainder of \(3\), \(x\) can be \(36+3=39\), \(72+3=75\), \(108+3=111\) ......... etc

The generalized form is \(36a+3 \) (where \(a\) is any positive integer)

\(x = 36a+3 \)

\(3(3k)^{\frac{1}{3}}= 36(a)+3\)

Cube both sides (remember \((a+b)^3 = a^3+3a^2b+3ab^2+b^3\))

\(3^4k = (36a)^3+3(36a)^2(3)+3(36a)(3)^2+3^3\)

\(k = \frac{(36a)^3}{3^4}+\frac{3(36a)^2(3)}{3^4}+\frac{3(36a)(3)^2}{3^4}+\frac{3^3}{3^4}\)

\(k =\) integer\(+\)integer\(+\)integer\(+\frac{1}{3}\)

As we can see, \(k\) is not an integer, though we know it to be an integer.

Therefore, there is no value of \(x\) where dividing \(x\) by \(36\) gives us a remainder of \(3\).

(3). \(27\) -> To get a remainder of \(27\), \(x\) can be \(36+27=63\), \(72+27=99\), \(108+27=135\) ......... etc

The generalized form is \(36a+27\) (where \(a\) is any positive integer)

\(x = 36a+27\)

\(3^3=27\)

\(3(3k)^{\frac{1}{3}}= 36(a)+3^3\)

Cube both sides

\(3^4k = (36a)^3+3(36a)^2(3^3)+3(36a)(3^3)^2+(3^3)^3\)

\(k = \frac{(36a)^3}{3^4}+\frac{3(36a)^2(3^3)}{3^4}+\frac{3(36a)(3^3)^2}{3^4}+\frac{(3^3)^3}{3^4}\)

\(k =\) integer\(+\)integer\(+\)integer\(+\)integer

As we can see, \(k\) is an integer.

Therefore, there exists a value of \(x\) where dividing \(x\) by \(36\) gives us a remainder of \(27\).

Answer - D ( I and III only)
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