12 Days of Christmas GMAT Competition - Day 2: For how many integer : Problem Solving (PS)

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SushantPaudel

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Re: 12 Days of Christmas GMAT Competition - Day 2: For how many integer [#permalink]

12 Dec 2023, 23:10

Here,
First, let's look at the prime factorization of 2,160 = (2^4)*(3^3)*5
So, When k=0 than 2160 *(2/3)^0 = 2160*1 = 2160
When k= 1 than 2160 * (2/3)^1 = (2^4)*(3^3)*5 * 2/3
Here, denominator 3 is cancelled by numerator 3.
We can get integer value of 2,160∗(2/3)^k when k's value are 0,1,2 and 3 because 3^3 in numerator cancel out denominator value.
So, 4 is answer i.e., B

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Re: 12 Days of Christmas GMAT Competition - Day 2: For how many integer [#permalink]

13 Dec 2023, 00:12

Let is first factorize this complex expression:

$2^33^3*(\frac{2}{3})^k$

now we can rewrite the expression as $2^(3+k)*3^(3-k)$

now since we need expression to be integer that means 2 or 3 can't be in denominator

hence,

$ 3+k\geq{0}$

and $ 3-k\geq{0}$

by solving above 2 we get $-3\leq{k}\leq{3}$

Hence possible integer values are -3,-2,-1,0,1,2,3. Total 7 integer possible. Answer is (C)

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Re: 12 Days of Christmas GMAT Competition - Day 2: For how many integer [#permalink]

13 Dec 2023, 01:16

The answer is A

2160 x 8/27 = is an integer

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Re: 12 Days of Christmas GMAT Competition - Day 2: For how many integer [#permalink]

13 Dec 2023, 01:21

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Bunuel wrote:

12 Days of Christmas 🎅 GMAT Competition with Lots of Questions & Fun

For how many integer values of k is $2,160*(\frac{2}{3})^k$ an integer?

A. 3
B. 4
C. 7
D. 8
E. 9

This question was provided by GMAT Club
for the 12 Days of Christmas Competition

Win $25,000 in prizes: Courses, Tests & more

when we factorize 2160 we get 3x3x3x2x2x2x2x1x5. hence, for this we can have k with the value of 0,1,2,3, -1,-2,-3,-4 and we will get integers. therefore the answer is 8 (D)

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Re: 12 Days of Christmas GMAT Competition - Day 2: For how many integer [#permalink]

13 Dec 2023, 01:47

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2160 = $2^4 * 3^3 * 5$

So, the expression will be integer for maximum of three 3s. Now, k = 0 also satisfies. For negative values of k, for maximum of four 2s the experssion will be integer. Hence total 8 values of k.

Ans: D

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Re: 12 Days of Christmas GMAT Competition - Day 2: For how many integer [#permalink]

13 Dec 2023, 01:48

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$2160 = 2^4*3^3*5$

For positive values of K expression will be $2^4*3^3*5*(\frac{2}{3})^k$
K can take values as 0,1,2,3

For negative value of k expression will be $2^4*3^3*5*(\frac{3}{2})^k$

K can take values as 0,-1,-2,-3,-4

Total values of k = -4, -3, -2 ,-1, 0, 1, 2, 3
Total 8 values
IMO Choice D.

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Re: 12 Days of Christmas GMAT Competition - Day 2: For how many integer [#permalink]

13 Dec 2023, 02:07

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Bunuel wrote:

12 Days of Christmas 🎅 GMAT Competition with Lots of Questions & Fun

For how many integer values of k is $2,160*(\frac{2}{3})^k$ an integer?

A. 3
B. 4
C. 7
D. 8
E. 9

This question was provided by GMAT Club
for the 12 Days of Christmas Competition

Win $25,000 in prizes: Courses, Tests & more

For how many integers value of k, $2,160*(\frac{2}{3})^k$ is also integer.
Now if we factorize 2160, we will get $2^4*3^3*5$

This term $2,160*(\frac{2}{3})^k$ will be integer considering two cases:
a. 2160 divisible by powers of 3 in case k is positive
b. 2160 divisible by powers of 2 in case k is negative

For k positive, we need divisibility by powers of 3:
So k can be 1,2,3

For k negative, we need divisitibility by powers of 2:
k can be -1,-2,-3,-4

k can also be zero

Thus k=8. Correct option is D

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Re: 12 Days of Christmas GMAT Competition - Day 2: For how many integer [#permalink]

13 Dec 2023, 02:19

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We need to check how many powers of 3 can perfectly divide 2160.
3,9, & 27 can divide perfectly.
Thus positive vales of k can be 1,2,&3.
If we consider negative values of k we have to check how many power of 2 can divide 2160.
2160 can be perfectly divided by 2 4 8 and 16
Thus k can -1,-2,-3,-4.
Taking k as 0 will also give an integer no.

Thus -4,-3,-2,-1,0,1,2,3
Therfore 8 values (answer D)

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Re: 12 Days of Christmas GMAT Competition - Day 2: For how many integer [#permalink]

13 Dec 2023, 02:27

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Bunuel wrote:

12 Days of Christmas 🎅 GMAT Competition with Lots of Questions & Fun

For how many integer values of k is $2,160*(\frac{2}{3})^k$ an integer?

A. 3
B. 4
C. 7
D. 8
E. 9

This question was provided by GMAT Club
for the 12 Days of Christmas Competition

Win $25,000 in prizes: Courses, Tests & more

$2160 = 3^3 * 2^4 * 5$

For $2,160*(\frac{2}{3})^k$ we can have the following values of 'k'

If k is positive then we have k = 1, 2, 3 as 2160 has the highest power of 3 as 3
If k is negative then we have k = 1, 2, 3, 4 as 2160 has the highest power of 2 as 4
If k is neither then we have k = 0

Total 8 options are available k can take 8 values
Option D

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Re: 12 Days of Christmas GMAT Competition - Day 2: For how many integer [#permalink]

13 Dec 2023, 03:53

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2160 =3^3 x 2^4 x 5

So for 3 power 0,1,2,3 The expression results in an integer.

And for k= -1,-2,-3,-4 the expression results in an integer. As 2 becomes denominator for negative sign of k.

This a total of 8 values of k result in an integer value of the expression.

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Re: 12 Days of Christmas GMAT Competition - Day 2: For how many integer [#permalink]

13 Dec 2023, 05:13

For how many integer values of k is 2160 *(2/3)^k an integer?

If we decompose 2160 in product of prime factors => 2160 = 2^4*3^3*5

2^4*3^3*5*(2/3)^k will be integer if and only if k<3 or k=3 so k can be 0, 1, 2 or 3 => so the answer is 4

so Answer B

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Re: 12 Days of Christmas GMAT Competition - Day 2: For how many integer [#permalink]

28 Mar 2024, 13:54

Bunuel wrote:

12 Days of Christmas 🎅 GMAT Competition with Lots of Questions & Fun

For how many integer values of k is $2,160*(\frac{2}{3})^k$ an integer?

A. 3
B. 4
C. 7
D. 8
E. 9

This question was provided by GMAT Club
for the 12 Days of Christmas Competition

Win $25,000 in prizes: Courses, Tests & more

2160=$2^4$*$3^3$*$5^1$
k can be positive, zero and negative.
When K is positive then 2160 has to be div by 3. From the prime factorization we can see power of 3 can maximum be 3 i.e. 0 to 3.
When K is negative then 2160 has to be div by 2. From the prime factorization we can see power of 2 can maximum be -4 i.e. -4 to 0.
Therefore, total number of integers are -4 to 3 i.e. 8. Option (D) is correct.

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Re: 12 Days of Christmas GMAT Competition - Day 2: For how many integer [#permalink]

28 Mar 2024, 13:54

GMAT Problem Solving (PS) Questions

12 Days of Christmas GMAT Competition - Day 2: For how many integer