If k is a positive integer and 6k^3 is divisible by 2500, then which

Question

If k is a positive integer and  6k^3  is divisible by 2500, then which of the following statements must be true?

I. k is divisible by 50
II. 20 is a factor of k
III.  k^5  is divisible by 64

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

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GMATinsight · Answer

6k^3 = 2500*x 
i.e.  2*3*k^3 = 2^2*5^4*x

i.e. k must be a multiple of 2 and 5^2 both

I. k is divisible by 50  - TRUE 
II. 20 is a factor of k  - Can't say because k may not be a multiple of 2^2 
III.  k^5  is divisible by 64  - Can't say because k may not be a multiple of 2^2

Answer: Option A

yashikaaggarwal · Answer

Sir, Can't K be 100 ? 
Because 6*100^3 is also divisible by 50 and has 20 as factor. Or there is some constraint?
Kindly revert.

Posted from my mobile device

GMATinsight · Answer

yashikaaggarwal 6k^3 = 2500*x i.e. 2*3*k^3 = 2^2*5^4*x i.e. 3*k^3 = 2*5^4*x Now the question asks "Which of the following MUST be true?" So we need to eliminate the options as much as possible. We can be certain about k to be a multiple of 2 but we can NOT guarantee that k is a multiple of 2^2 e.g. if x = 3*2^2*5^2 then k_{min} = 2*5^2 i.e. k is NOT necessarily a multiple of 20 akd k^5 is not divisible by 64 If the language were which of the following COULD be true? then we must have done what you are suggesting. I hope this help!

ScottTargetTestPrep · Answer

Solution:

Since 6 = 2 x 3 and 2500 = 25 x 100 = 5^2 x 2^2 x 5^2 = 2^2 x 5^4, k has to have at least one factor of 2 and at least two factors 5 (and hence k^3 has three factors of 2 and six factors of 5) so that 6k^3 is divisible by 2500. Now, let’s check the statements.

I. k is divisible by 50

Since k has a factor of 2 and two factors 5 and 2 x 5^2 = 50, statement I is true.

II. 20 is a factor of k

We see that if k = 50, statement II will not be true.

III. k^5 is divisible by 64.

If k has only one factor of 2, then k^5 is visible by 2^5 = 32, but not by 64. Statement III is not true.

Answer: A

HoneyLemon · Answer

If k is a positive integer and 6k^3is divisible by 2500, then which of the following statements must be true?

6k^3 is divisible by 2500

K must be 50n .

I. k is divisible by 50  -- TRUE  
II. 20 is a factor of k  -False  
III. k^5 is divisible by 64  -False

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

Ans is  A

Lipun · Answer

Hi  Bunuel ,

I think the OA needs correction.

Given, 6k^3 = 2500*x
=> 2*3*k^3 = 2^2*5^4*x
=> 3*k^3 = 2*5^4*x
=> 2*5^4 is a factor of k^3.

Since all the prime factors in the cube of a number will have power in the multiples of 3, so k^3 will have a minimum of 2^3 and 5^6 in it's prime factorization.

Let k^3 = 2^3*5^6*p
=> k = 2*5^2*q

From the above we can't be sure that k will be a multiple of 20 as 'q' may or mayn't be a multiple of 2.

Thanks
Lipun

chetan2u · Answer

6k^3  is divisible by 2500 or  5^4*2^2  =>  6k^3=5^4*2^2*x.......3k^3=5^3*5^1*2^1*x 
So k has to have single power of each of the term in expansion  5^3*5^1*2^1 , so  5*5*2*x=50x 
So k=50x, where x is a positive integer.

I. k is divisible by 50
 As k=50x, the statement is always true.

II. 20 is a factor of k
 As k=50x, 
If x is odd, the statement is false. But, if x is even, the statement is true.

III.  k^5  is divisible by 64
 k=50x=2*5^2*x....k^5=2^5*5^{10}*x^5=32*5^{10}*x^5

If x is odd, the statement is false. But, if x is even, the statement is true.

Only I is 'Must be true', so A.

NamakUoPada · Answer

Since X=6k^3 is divisible by 2500, X should contain all factors of 2500.

Since 6 is not a factor of 2500, X is divisible by 2500 only because of k^3.

After applying prime factorization in 2500, we have

k^3=L*( 2^2)*( 5^4) with L being an integer.

In other words, k^3 should contain 2 two times and 5 four times. In order to achieve that, k must contain 2 once and 5 twice. 
So it must be true only that k = n50, (with n being an integer).

tejasvkalra · Answer

6k^3/2500= an integer 
2*3*k^3/5^4 * 2^2 = integer

3*k^3/5^4 * 2 = integer 
Thus , k must have atleast 5^2 and 2^1

so minimum value of k= 5^2 * 2 = 50
I always has to be true
II , 20 has to be a factor of k....not necessary 
III k^5 is divisible by 64
for our minimum value of k , we have k^5 = 5^10 * 2^5.....this is not divisible by 64

Thus only I has to be true

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If k is a positive integer and 6k^3 is divisible by 2500, then which

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If k is a positive integer and 6k^3 is divisible by 2500, then which

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