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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

Sketchitesh

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07 Jul 2025, 12:18

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x = positive integer.
x^3 = 81n (for some positive integer n)

x^3 = 3^4 * n => 3^3 * 3n => n must be at least 3^2 = n for x to be postive integer.
=> x= 9m (for some positive integer m)

Q: remainder when x/36 = x/ (9*4)

I. 0
9m/36 => for all m multiple of 4, there will be 0 as remainder.

II. 3

for 3 as remainder x= 36a+ 3, for some integer a.
=> 36a+3 = 9m => 4a + 1/3 = m . Since a is integer, m can't be integer. no possible case of 3 as remainder.

III. 27
similarly 36a+27 = 9m => 4a+3=m . Possible for any positive integer a.

I and III only. 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

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07 Jul 2025, 12:23

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Bunuel

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81 = 9 * 9 = 3^4

x is a multiple of 9

x = 9*p for some constant p

Remainder (9*p / 36) = 9 * Remainder (p/4)

When a number is divided by 4, the remainder can be 0, 1, 2, 3

In this case the possible remainders are (0*9), (1*9), (2 * 9) , (3*9)

I. 0 - VALID
II. 3 - NOT VALID
III. 27 - VALID

Option D - I and III only

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07 Jul 2025, 12:41

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$81=3^4$
$x^3=3^4*A$
For $x^3$ to be a multiple of $3^4$, x needs to be a multiple of $3^2$.
So we can write x=9k, where k is a positive integer.
Now we want to find the remainder when x is divided by 36.
$9k=36q+r$, so : $r=9k-36q=9(k-4q)$
So the remainder r also needs to be a multiple of 9, which means 0 and 27 could be the remainder, but not 3.

Option D is correct.

Bunuel

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07 Jul 2025, 12:48

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a multiple "of by 81" is confusing" but either way, it seems to be 81 multiplied by something. 81 is multiple of 3, and therefore, whatever it's multiplied by must be a multiple of 3 per the rule that if x does not divide N, then x * y will not divide N.

we know that 3 divides N, because 81's digits adds up to a number that is divisible by 3: i.e., 8+1 = 9/3 = 3; or simply put 81/3= 27

since 3 divides 81. 81 x any integer will be divisible by 3 --> even 81 time an even integer like 4.

81 x any number will provide a multiple whose digits add up to a number that is divisible by 3

e.g., 81 x 6 = 486 divided by 3 = 162

36= a number whose digits add up to 9; i.e., 3+6 = 9

a multiple of 3/ divided by a multiple of 3 will always have a remainder of zero. --> answer: I only

Bunuel

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07 Jul 2025, 12:55

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Given:
$x^3$ is a multiple of by 81= $3^4$
So $x^3$ must contain at least four 3s in its prime factorization
i.e, x must have at least (4/3) = 2 powers of 3
==> x is divisible by $3^2$= 9

So, x can be written as 9n(where n is some positive integer)
==> x= 9n

Possible remainders when x= 9n is divided by 36

1) n=1, x= 9 Remainder when 9 is divided by 36 =9
2) n=2, x=18 Remainder when 18 is divided by 36 =18
3) n=3, x=27 Remainder when 27 is divided by 36 = 27
4) n=4, x=36 Remainder when 36 is divided by 36 = 0
.....

From the given options I, III are possible Option D

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07 Jul 2025, 13:10

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Bunuel

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Given that x >0 a positive number.

And x^3 is a multiple of 81. For x^3 to be a multiple of 81. We need x be a multiple of 9. As x = 3 will yield 27 on x^3.

Thus, x = 9K.

What would be the remainder when x divided by 36.

9k/36 remainder can be 0 if k=0

Remainder = 9, if K=1

Remainder = 18, if k=2

Remainder = 27, if k = 3

Remainder = 0, if K =4

So the possible values of remainder are { 0, 9, 18, 27}

Possible options are I and III.

So, Option D.

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07 Jul 2025, 13:35

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Topic(s)- Remainders, Factors, Cubic Roots
Strategy- Algebra, Intuition
Variable(s)- # of multiples = "K," "L," "M," "N"

1. "English" -> Math
i) x times x times x is some number of 81's
x^3 = 81*K
ii) x times x times x is some other number of 3's
x^3 = (3^4)*L
iii) For x to be a positive integer, x^3 needs to at least be some other number of (3^6)'s and not (3^4)'s
x = integer =\= cubic root (3^4)
x = integer = cubic root ((3^6)*M)) = (3^2)*N
x = some number of 9's

2. Remainder Pattern for x/36 = (9*N)/36
i) Given N = 1, 2, 3, ...
R(N=1) = 9*(1)/36 = 9/36 -> Remainder = 9
R(N=2) = (18/36) = 18/36 -> Remainder = 18
R(N=3) = (27/36) = (3/4) -> Remainder = 27
R(N=4) = (36/36) -> Remainder = 0
R(N=5) = (45/36) = (36/36 + 9/36) = (1 + 9/36) -> Remainder = 9
ii) R = {9, 18, 27, 0, repeat...}

Answer: D

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07 Jul 2025, 13:37

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Ans (D)

Given that x^3 = 81

Possible values of x = 9, 27, 30, 33, 33, 36, 39

However, the remainders would be 0, 9, 18, 27, ....

I x/36=0

Possible since x=36 is a possible value as 36 is a multiple of 9

II x/36=3

Not possible since 39 is not a multiple of 9

III x/36=27

Possible since 27 is a multiple of 9

Bunuel

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07 Jul 2025, 14:15

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x>0 and x^3 is a multiple of 81.
81 = 3^4
This implies x^3 has powers of 3 >= 4 which start from 3^6 since it is a cube.
36 = 2^2 * 3^2

Case I
If x = 36, x^3 % 81 = 0 and x % 36 = 0.
POSSIBLE

Case III
If x = 27 = 3^3, x^3 = 3^9. 27^3%81 = 0 and 27%36 = 27.
POSSIBLE

Case II
If x%36 = 3, x = ky + 3
It is given that x^3%3^4 = 0. This implies (3(ky + 1))^3 %81 = 0. This situation cannot happen since power of 3 will be odd.

Answer is D i.e. I and III only.

Bunuel

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07 Jul 2025, 14:49

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Bunuel

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Very fast solvable trough a basic testing approach.
We have muliples of 81 and need to find a pattern in the division of 36.
Of course, we could use the fact, that we are talking about 3^4 = 81 and 2^2*3^2

But i chose to go with pattern recognition.
81/36 = 2; remainder: 9
2*81/36 = Remainder: 18
3*81/36 = Remainder: 27
4*81/36 = Remainder: 0
and the same again.

Therefore, only I and III are applicable.

Answer is D)

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The correct answer is (A) I Only

If we know that X^3 is a multiple of 81, then we need to check out the factors of 81 to see if X^3 would be divisible by them. That would inform us more about possible remainders.

81 can be broken down into a few factors - 81x1, 9x9, 27x3, 3x3x3x3.

Of the roman numeral options, both 27 and 3 are factors of 81 so they could divide evenly into x^3 and not leave a remainder. Therefore the only option possible is I Only, a remainder of 0. This gives us answer choice A.

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07 Jul 2025, 17:07

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Bunuel

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This is a interesting question. When we think about xcube being multiple of 81. We can write "x = 9 x (N/9)_1/3". But since in question it is already cleared that x is a positive integer which means " (N/9)_1/3" is a integer aswell. Now we dont need to worry about " (N/9)_1/3" and focus on " 9 x (N/9)_1" and write it as, 9K where K is any integer. When we divide 9K by 36,

it is clear that remainder would be either 0 when K is any multiple of 4,
remainder will be 9 when K is any number starting 1 with a difference of 4 , like 1, 5, 9, 13, ...., etc.
remainder will be 18 when K is any number starting 2 with a difference of 4, like 2, 6, 10, ... etc.
Lastly, it will have remainder of 27 when K is number starting 3 and have difference of 4, like 3, 7, 11, 15, ... etc.

So we can clearly see, remainder is either 0, 9, 18 or 27. So correct answer is (D)

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07 Jul 2025, 18:18

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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?

x >=1 & x^3 = a * 81 = a * 9 * 9 = a * 9^2 = a * 3^4; To make this cube we need the supporting integers.

Some possible values of x^3:
x^3 = 9*9*9; X^3 = 3^6; X^3 = 3^3m * 2^3n; (but m should be greater than 1, for it to be a multiple of 81)

x = 3^m * 2^n

Now, 36 = 9*4 = 3^2 * 2^2;

x/36 = remainder + (x^3)^-3

I. 0 -> This could be a remainder;

II. 3 -> By calculating for multiple samples, you will find that this can't be a remainder.

III. 27 -> for x = 3^3 = 27; x^3 = 3^9; Multiple of 81; Means that this can be a remainder

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

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07 Jul 2025, 18:48

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Bunuel

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We know that x^3 is 81k;

x^3 = 3^4. k => So it's obivious that k has to be 3^2y. (Since x is a +ve integer) .

Hence x = 3^2t. (t is a +ve int).

Now x/36 => 9t/36 => So we can take 9 common, Which means this can take any rmider from 9,18,27,0. (Since we are dividing by 4 after bringing 9 common).

Hence 0,27 are possible.

IMO D

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07 Jul 2025, 18:53

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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?

I. 0
II. 3
III. 27

Let x= 36
X3= 36*36*36= 81*9*64
Clearly x3 is divisible by 81. X is divisible by 36
Option I holds true as remained zero.

Let x= 3
X3= 27 not divisible by 27

x should be min 9 for X3 to be divisible by 81. Thus remainder 9 is poosible with 36 like 9,45,81 etc
X=3 doesn't hold true.

Let x= 27
X3= 27*27*27
Clearly x3 is divisble by 81
X/36 has remainder 27
Same is possible for 63 likewise.
So option III holds true

Answer should be I and III only i.e. D

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07 Jul 2025, 19:51

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81=3^4
if x^3 is a multiple of 81, x must have at least 3^2 in its prime factorization (if x has only only one 3, in x^3 there would be 3^3=27, that is not a multiple of 81).
So x is divisible by 9=3^2.

I. x = 36q + 0. It is possible that x has this form because 36 is divisible by 9.
II. x = 36q + 3. It is impossible that x has this form because 36 is divisible by 9 and 3 isn't divisible by 9.
III. x = 36q + 27. It is possible that x has this form because 36 is divisible by 9 and 27 is divisible by 9.

I and III are possible

IMO D

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07 Jul 2025, 19:52

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D. I and III

81 = 3 * 3 * 3 * 3
36 = 3 * 3 * 2 * 2

You need at least 9 factor.

I.
x^3 needs to have more factors of 3 than both.
36^3 has 3^6 -> works.

II.
Trying 3: 3^3 -> not enough 3 factors
Trying 39: 39^3 = 3^3 * 13^3 -> not enough 3 factors
Trying 75: 75^3 = 3^3 * 25^3 -> not eough 3 factors
-> 3 factors stay the same, hence, DOES NOT WORK

III.
27= 3^3
27^3 = 3^9 (divisible by 81)
27/36 makes this work.

Bunuel

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07 Jul 2025, 20:08

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$x^3=81n=3^4n=3^3*3n$

$n$ must be a multiple of $3^2or9$; $n=3^2k^3$

$x^3=3^3*3^3 k^3$

$x=9k$

$x$ is a multiple of 9

When $x=36k+0$, remainder is 0 when divided by 36. $x$ is divisible by 9 & $x^3$ is divisible by 81.

when $x=36k+3$, remainder is 3 when divided by 36. $x$ is not divisible by 9. Could not be the remainder

when $x=36k+27$, remainder is 27 when divided by 36. $x$ is divisible by 9 & $x^3$ is divisible by 81.

Answer: D

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07 Jul 2025, 20:47

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I. Remainder = 0:
If x leaves remainder 0 when divided by 36, it means 36 divides x. Since 36 includes 3^2 = 9, x is divisible by 9, so x^3 is divisible by 81.

II. Remainder = 3:
If x leaves remainder 3 when divided by 36, then x is not divisible by 9. Without a factor of 9, x^3 cannot have enough factors of 3 to be divisible by 81.

III. Remainder = 27:
If x leaves remainder 27 when divided by 36, then x contains a factor of 9. Therefore, x^3 has enough factors of 3 to be divisible by 81.

Answer : D (1 and 3 only).

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07 Jul 2025, 21:14

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x3 is a multiple of 81(3^4)
x should be divisible by at least by 9 since x should be a multiple of 3^4/3*a^1/3.( a should be min 9)

option 1: 0
number= 36K+0=36k divisible by 9 so it is correct

option 2:3
number= 36k+3 not divisible by 9 so not correct

option 3: 27
number= 36K+27 divisible by 9 so it is correct

therefore option 1 and 3 are correct.