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

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Bunuel

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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

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

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Bunuel

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Bunuel

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

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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

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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).

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Bunuel

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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.

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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

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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

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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?

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

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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

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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.

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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

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Bunuel

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Let x be 27 therefore x^3 is 27*27*27/36=27*27*3/4 hence 3 is the remainder

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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.

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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

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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)

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Bunuel

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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

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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

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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

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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