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GMAT Club Olympics 2025 (Day 6): If x is a positive integer and x^3 is

1 2 3 4 5

Prathu1221

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For x^3 to be multiple of 81 x should have atleast 2 3s means minimum 9 should be there then only x^2 would be divisible by 81.
If we write as 9m for value of x^3 then it is a cyclic remainer of 9, 18 27,0 and 9 again.
So option D is right.

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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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 $36$?

Given $x^3$ is a multiple of $81$
We have $x^3 = 81y = (3^4)y$

This mean $x^3$ should have a minimum of product of four 3's; 3*3*3*3 = 81
For this to be true, $x$ must be a multiple of $9$; $x = 9a$
Now considering the multiples of $9$ and its remainders,

$\\
9 - 9/36 -> R = 9\\
18 - 18/36 -> R = 18\\
27 - 27/36 -> R = 27\\
36 - 36/36 -> R = 0\\
45 - 45/36 -> R = 9\\
54 - 54/36 -> R = 18\\
63 - 63/36 -> R = 27\\
$

We can see the remainders possible as 9, 18, 27 & 0; The options I & III are the only ones correct.

So the correct option is Option D

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Analyze the constraint
Since x3 is a multiple of 81 = 34, we need x3 to be divisible by 34.
For x3 to be divisible by 34, x must be divisible by at least 32 = 9. (Since if x = 3a × k where gcd(k,3) = 1, then x3 = 33a × k3, and we need 3a ≥ 4, so a ≥ 4/3, meaning a ≥ 2)
So x must be a multiple of 9.

Check each option

Option I: Remainder 0 If x ≡ 0 (mod 36), then x is a multiple of 36. Since 36 = 9 × 4, x is also a multiple of 9.

Option II: Remainder 3 If x ≡ 3 (mod 36), then x = 36k + 3 = 3(12k + 1). Since 12k + 1 is not divisible by 3, x is only divisible by 31, not 32. So x is not a multiple of 9.

Option III: Remainder 27 If x ≡ 27 (mod 36), then x = 36k + 27 = 9(4k + 3). So x is a multiple of 9. ✓

Verify with examples

For remainder 0: x = 36, x3 = 46,656 = 81 × 576
For remainder 27: x = 27, x3 = 19,683 = 81 × 243

Answer: D. I and III only

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Correct Answer: Option D (I and III only)
Let’s check possible values of x^3 which can be divisible by 81,
1) We can keep x=9 as, 9^3= 729 which is divisible by 81.
2) We can keep x= 27 as, 27^3= 19683, which is also divisible by 81.
Now using same approach we will find out what possible remainders we get when x is divided by 36:
1) 0
If we take x= 36 which includes factor as 9*4 so we can definitely say that 36^3 is divisible by 81 as well because 9^3 is divisible by 81. This is correct.
2) 3
This cannot be correct 3^3 is 27 for divisible by 81 the power of 3 must be greater than 3 so this option is incorrect.
3) 27
If we keep x=27 as, 27^3= 19683, which is also divisible by 81.
So if we divide 27 by 36 then the remainder we have is 27.
Thus this option is also correct.
So both I and III are correct.

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Bunuel

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x>0 and x^3 is a multiple of 81.
81= 3^4
=> For a number to be divisible by 81, it should at least have 4 factors of 3.

But x is a cube root.
=> x should have at least 2 factors of 3 i.e., x will always be divisible by 9.

I. 0
Yes, it can be a remainder.
If x=9
Then, 36/9 will have remainder=0.

II. 3
Here, x=36k+3
Which will only have one factor of 3 but, these won't make x^3 a multiple of 81.
Moreover, x is always divisible by 9.
No.

III. 27
x= 27
36/27 will give remainder 27.
Yes.

Answer: D I and III only.

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D. I and III only

Going through I,II,III respectively

I. 0
which means x can be written as
x = 36y
we can substitute y with any positive integer so that x^3 is a multiple of by 81, e.g. y=1,
so I is possible

II. 3
similar to I.
x = 36y + 3
writing x^3
x^3 = (36y)^3 + 3^3 + 3(36y)(3)(36y + 3)
A B C
A,C is divisible by 81 but B is not, for the whole thing to be a multiple of 81, B has to be a multiple of 81, which means II. can not be the remainder

III. 27
similar to II.
x = 36y + 27
writing x^3
x^3 = (36y)^3 + 27^3 + 3(36y)(27)(36y + 27)
A B C
Here, all A,B and C are divisible by 81, so III. can be the remainder

thus answer is D. I and III only

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x^3 is a multiple of by 81 = 3^4 x k
so k must have atleast 9 as its multiple so that x becomes an integer
so x = 9n
hence x can be 9,18,27,36,45,54,63 and so on

now when divided by 36, what can be the remainder:

0: when n=4 the remainder will be zero
3: not possible as x cannot be 39 or 75 or 111.
27: when x = 63, the remainder will be 27.

So I and III are possible

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the answer to the question is B.

Q given:
x^3/81 = X is ODD
so 0 cannot be REMAINDER
we CAN ELIMAMTE A,D and E- On this basis

36=3*3*2*2 which means 27 cannot come as are remainder- so the answer is B

Bunuel

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$x^3$ = 81k = $3^4k$

so x must be a multiple of n since there must be a 3^2 factor in the factorization of x for 3^4 to be able to divide n
let x = 9n
Also, 36 = 9 * 4

for n=1, x =9, when divided by 36, will give remainder = 9
for n=2, x =18, when divided by 36, will give remainder = 18
for n=3, x =27, when divided by 36, will give remainder = 27
for n=4, x =36, when divided by 36, will give remainder = 0

so both 27 and 0 are possible, but 3 is not possible.
Answer is C

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1. Remainder =0 Yes if x is a multiple of 36
2. x=3 No because 3^3 not divisible by 81
3. x=27 yes cuz it s cube is divisible by 81

Ans D

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x>0 , x^3 multiple of 81=3^4.
x^3 is divisible by 3^4
This gives x must be divisible by 3^(4/3)=3^2=9
I. 0
II. 3
III. 27
1.remainder=0, this means x is divisible by 36, 36 is divisible by 9, so true.
2. remainder=3, this means x=3 or 39, none of them are divisible by 9
3.remainder=27, this means x=27 or 63, divisible by 36.
IMO:D

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Bunuel

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x^3 is a multiple of by 81
=> x^3 = 81k
=> x^3 = 3^4 * k
=> x^3 = 3^4 * 3^2 * m [Since x must have at least 3^6]
=> x^3 = 3^6 * k
=> x = 3^2 * k^(1/3)
=> x = 9a

Now if 9a is divider by 36 then we have
9 / 36 --> Remainder 9
18 / 36 --> Remainder 18
27 / 36 --> Remainder 27
36 / 36 --> Remainder 0
45 / 36 --> Remainder 9
54 / 36 --> Remainder 18
63 / 36 --> Remainder 27

This is a pattern hence we can find which numbers are satisfying
I. 0 --> Correct
II. 3 --> No
III. 27 --> Correct

D. I and III only

Option D

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For x^3 to be a multiple of 81, x must be divisible by 9 ( because 81 = 3^4 and taking the cube root shows x needs >=2 factors of 3).
When any multiple of 9 is divided by 36, the remainder can only be 0, 9, 18, or 27.
so only 0 and 27 are feasible, not 3
Answer : D

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As x^3 is a multiple of 81, then x must be equal to 3^2 multiplied by some integer because (3^2)^3 = 3^6, what guarantees that x^3 is divisible by 81=3^4.
x is a multiple of 9.

x = 36*quotient + remainder
As x is a multiple of 9, and 36*quotient is divisible by 9, all the possible remainders must be divisible by 9.

I. 0 -> divisible by 9
II. 3 -> not divisible by 9
III. 27 -> divisible by 9

I and III only

The right answer is D

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Ans: D

Given x^3 = 81K = (3^3) * 3K
Since above expression is a cube root then 3 must have powers divisible by 3 => the power of 3 must be at-least 6 (currently its 4 in the above exp)
=> x = 3*3K' =9K'
Rem(9K'/36) where K' is a +ve integer as x is a +ve integer
For K'=1 Rem = 9
K'=2 Rem= 18
K'=3 Rem= 27
K'=4 Rem=0
So we see the remainder are multiple of 9 hence 0 & 27 can be remainder when x is divided by 36.

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I did this by hit and trial. Because x^3 is multiple of 81, it can be something with triplets as factors and multiple of 81. Tried x^9=729 hence x is 9, when divided by 36, remainder comes as 9, which isnt given to us in any of te options, then went on with x^9 , x=27, divided by 36, 27 is the remainder, 0 and 3, cant be remainders , fulfilling these conditions. So, C

Bunuel

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Given,
x>0
x3 = 81k
Cube-root both sides
∛x3 = ∛81k
x = ∛34k
k has to be at least 32z as x is a +ve int and z2(P.F. of 34z other than 3)
x = ∛36z2
x = 32z = 9z
(9z/36)R

If z=1, R=9
If z=2, R=18
If z=3, R=27
If z=4, R=0
So, only 3 cannot be a remainder

Option D

Bunuel

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Given that 81=3^4
36 = 2^2X3^2
X^3= 3^4XA
I = (3^4XA)/ 2^2X3^2 =3^2XA/2^2. A could be divisible by 2^2 meaning the remainder can be 0
II= (3^4XA)/ 2^2X3^2= 3^2XA/2^2 = 9A/4 Not possible to have 3 as a remainder
III= 9A/4 = rem 27 very possible assuming A is 3 then 9x3/36 is rem 27
Hence I and III

Bunuel

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x^3= a number which is a multiple of 81.

When x^3 is divided by 36 what remainder does it give

81= 3^4
36=3^2* 2^2

Multiples of 81

1*81
3*27
9*9

Multiples of 36

1*36
2*18
3*12
4*9
6*6

A number x^3 that is divided by 36 and would leave a reminder

Is a multiple of 36+R and this value should be divisible by 81.

R is the remainder and the range is between 0 and 36

Meanwhile 81 is 9^2

Option 1

R=0
The cube root of all the larger multiples of 36:36^3, 18^3,12^3, 9^3 are divisible by 81 because the have 9^2 in their factors

Hence option 1 is correct

Option 2

R= 3

36^3+3 ,18^3+3, 12^3+3,9^3+3 are not multiples of 81. The remainder 3 is not divisible by 9

So this option is not correct

Option 3:

R=27

The remainder is divisible by 9.hence it is a multiple of 81.

So option 3 is correct.

Hence the answer is d

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Given:
x^3 is positive integer and is multiple of 81.

So, for above to be true we first have to see prime factors of 81, i.e, 3^4

So if we see answer choices,

1. 0, 36 needs to x, lets see the factors: 3^2*2^2 so if we cube 36 we can have 3^4 so this can be one of the solutions so remainder 0 is possible.

2.3, For 3 to be a remainder min value for x will be 39 which means 3*13, so if we cube this we will not get 81 as factor so no remainder 3 is not possible.

3. 27, for 27 to be remainder min value of x will be 27 which means 3^3 is the prime factorisation, if we cube it it will become factor of 81 so, this is possible.

Answer: D. I and III only

Bunuel

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