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If k and x are positive integers and x is divisible by 6 [#permalink]

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29 Oct 2010, 17:10

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If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of \(\sqrt{288kx}\) ?

A) \(24k\sqrt{3}\) B)\(24\sqrt{k}\) C)\(24\sqrt{3k}\) D)\(24\sqrt{6k}\) E)\(72\sqrt{k}\)

I got it wrong and I have had a look to the explanation and I am just wondering if it can be solved differently than the official explanation of MGMAT. I also thought it could be interesting as I have not found a previous thread with this question.

If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of \sqrt{288kx} ?

A) 24k\sqrt{3} B)24\sqrt{k} C)24\sqrt{3k} D)24\sqrt{6k} E)72\sqrt{k}

I got it wrong and I have had a look to the explanation and I am just wondering if it can be solved differently than the official explanation of MGMAT.

x is divisible by 6 --> \(x=6n\) for some positive integer \(n\).

Now, for any values of positive integers \(k\) and \(n\) \(24\sqrt{3kn}\) is always more than \(24\sqrt{k}\) (B): \(24\sqrt{3kn}>24\sqrt{k}\) --> \(\sqrt{3n}>1\).

Thanks. It is kind of the same approach of the one in MGMAT so I guess I am gonna have to stick to this. I guess I don't have the initiative to say x=6n and then to compare.

This is not an ez one... u can also put x=6, k=1 and you can see A,B,C goes out quickly bc u found a very ez to spot solution for them... _________________

How can one be sure that the expression here is 288*k*x and not 28800 + 10*K + x where K and x are integers?

Only multiplication sign is usually omitted so 288kx means 288*k*x. If it were meant to be 5-digit integer 288kx, where k is a tens digit and x is a units digit then this would be explicitly mentioned in the stem. _________________

I still dont understand it. It is clear that 24\sqrt{3kn} > 24\sqrt{k}. But isn't it also greater than 24\sqrt{3k} ?

Not really. For, x=6, n=1; \(24\sqrt{3kn}=24\sqrt{3k}\)

Likewise; For, n=k; \(24\sqrt{3kn}=24\sqrt{3k^2}=24k\sqrt{3}\) For, x=12, n=2; \(24\sqrt{3kn}=24\sqrt{6k}\) For, x=18, n=3; \(24\sqrt{3kn}=24\sqrt{3^2k}=72\sqrt{k}\)

But to make \(24\sqrt{3kn}=24\sqrt{k}\), n should be 1/3 and we know that n is an integer. Not possible. _________________

Option a- Posible if n*x=k^2 Option b- Not possible as x is an integer, it cannot be a fraction. (6n=k=an integer, x is an integer). For this to be true, k=1 and x=2 not possible as k is a multiple of 6 or k=3 again not possible.

Re: If k and x are positive integers and x is divisible by 6 [#permalink]

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26 Sep 2013, 13:02

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If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of \sqrt{288kx} ?

A) 24k\sqrt{3} B)24\sqrt{k} C)24\sqrt{3k} D)24\sqrt{6k} E)72\sqrt{k}

I got it wrong and I have had a look to the explanation and I am just wondering if it can be solved differently than the official explanation of MGMAT.

x is divisible by 6 --> \(x=6n\) for some positive integer \(n\).

Now, for any values of positive integers \(k\) and \(n\) \(24\sqrt{3kn}\) is always more than \(24\sqrt{k}\) (B): \(24\sqrt{3kn}>24\sqrt{k}\) --> \(\sqrt{3n}>1\).

Answer: B.

Can you explain how you know that? I got C for an answer...

If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of \sqrt{288kx} ?

A) 24k\sqrt{3} B)24\sqrt{k} C)24\sqrt{3k} D)24\sqrt{6k} E)72\sqrt{k}

I got it wrong and I have had a look to the explanation and I am just wondering if it can be solved differently than the official explanation of MGMAT.

x is divisible by 6 --> \(x=6n\) for some positive integer \(n\).

Now, for any values of positive integers \(k\) and \(n\) \(24\sqrt{3kn}\) is always more than \(24\sqrt{k}\) (B): \(24\sqrt{3kn}>24\sqrt{k}\) --> \(\sqrt{3n}>1\).

Answer: B.

Can you explain how you know that? I got C for an answer...

That's because for any values of positive integers \(k\) and \(n\), \(\sqrt{3kn}>\sqrt{k}\) _________________

If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of \sqrt{288kx} ?

A) 24k\sqrt{3} B)24\sqrt{k} C)24\sqrt{3k} D)24\sqrt{6k} E)72\sqrt{k}

I got it wrong and I have had a look to the explanation and I am just wondering if it can be solved differently than the official explanation of MGMAT.

x is divisible by 6 --> \(x=6n\) for some positive integer \(n\).

Now, for any values of positive integers \(k\) and \(n\) \(24\sqrt{3kn}\) is always more than \(24\sqrt{k}\) (B): \(24\sqrt{3kn}>24\sqrt{k}\) --> \(\sqrt{3n}>1\).

Answer: B.

Makes complete sense when the steps are listed out in such a way. Question though -- when I read the statement that x is divisible by 6, I started to work with the primes (2 & 3) and tried to correlate that to the prime of (288). Why is that approach incorrect in this problem?

I realize that there is no sure fire way to encompass every problem but it just seems odd that we solve a lot of problems by factoring down to their primes but in this case, we added an unknown variable to solve the problem?

If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of \sqrt{288kx} ?

A) 24k\sqrt{3} B)24\sqrt{k} C)24\sqrt{3k} D)24\sqrt{6k} E)72\sqrt{k}

I got it wrong and I have had a look to the explanation and I am just wondering if it can be solved differently than the official explanation of MGMAT.

x is divisible by 6 --> \(x=6n\) for some positive integer \(n\).

Now, for any values of positive integers \(k\) and \(n\) \(24\sqrt{3kn}\) is always more than \(24\sqrt{k}\) (B): \(24\sqrt{3kn}>24\sqrt{k}\) --> \(\sqrt{3n}>1\).

Answer: B.

Makes complete sense when the steps are listed out in such a way. Question though -- when I read the statement that x is divisible by 6, I started to work with the primes (2 & 3) and tried to correlate that to the prime of (288). Why is that approach incorrect in this problem?

I realize that there is no sure fire way to encompass every problem but it just seems odd that we solve a lot of problems by factoring down to their primes but in this case, we added an unknown variable to solve the problem?

Thanks!

We are told that a positive integer x is divisible by 6. It can be 6, 12, 18, 24, ... So, it can have some other primes than 2 and 3. That's why we need to represent it as 6n, where n is a positive integer and not just assume that it's 6.

As for factorization: as you can see that along the way we do factorize 6n into 2*3*n.

Re: If k and x are positive integers and x is divisible by 6 [#permalink]

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03 May 2015, 06:46

I am really bad with number properties... So what I did was estimating the squareroot of 288XX which I thought to be something around 170. Then I went through the solutions estimating which of them could be around 170, and I could elimate B. 24 times squareroot of 9 ( the highest digit for k) would def be less than 170

Re: If k and x are positive integers and x is divisible by 6 [#permalink]

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04 May 2015, 03:50

Expert's post

ZLukeZ wrote:

I am really bad with number properties... So what I did was estimating the squareroot of 288XX which I thought to be something around 170. Then I went through the solutions estimating which of them could be around 170, and I could elimate B. 24 times squareroot of 9 ( the highest digit for k) would def be less than 170

This is definitely not the way to solve this question. Even though you got the right answer, let me tell you the gaps in your methodology:

1. Please note that k and x are not single unit digits at the tens and units place of the number 288kx. The number is 288 *k *x. Hence, the number is not around 288XX for which the square root will be around 170. x can be any multiple of 6 and k can be any other positive integer. So value of 288kx can be far greater than 288XX.

2. Since k and x may not be single unit digits, assuming maximum value of k to be 9 is not correct. k is a positive integer and can take any value which may be greater than 9.

Simplifying the number by prime factorizing and then evaluating the options would be the right way to go about this question.

Hope its clear! Let me know if you need help at any point of this question.

Re: If k and x are positive integers and x is divisible by 6 [#permalink]

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27 Jan 2016, 15:07

Could someone please tell me if this is a correct way to solve the problem.

Given that X is divisible by 6, it has prime factors of 3 and 2. Prime factors of 288 are 2,2,2,2,2,3,3 Combined, under the square root, we have the prime factors 2,2,2,2,2,2,3,3,3, and k

I concluded that B is the answer because it was missing a prime factor 3.

Re: If k and x are positive integers and x is divisible by 6 [#permalink]

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27 Jan 2016, 18:24

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Expert's post

ZaydenBond wrote:

Could someone please tell me if this is a correct way to solve the problem.

Given that X is divisible by 6, it has prime factors of 3 and 2. Prime factors of 288 are 2,2,2,2,2,3,3 Combined, under the square root, we have the prime factors 2,2,2,2,2,2,3,3,3, and k

I concluded that B is the answer because it was missing a prime factor 3.

Yes, that is a correct approach but an incomplete one. When x = 6p, where p=positive integer, then x can be any 1 of multiples of 6 = 6 or 12 or 18 or 24 etc. So you would have prime factors (2*3*p) and not just (2*3). This is a very small but important difference that might become important in other questions.

Another approach to this question is as follows:

We are given that x is a divisible by 6 ---> x = 6p =2*3*p

Now analyse the options (a very important tool in GMAT quant) to eliminate options that are possible (as we need to mark the option that is NOT possible).

\(A) 24k\sqrt{3}\). Possible with p=k. Eliminate \(B)24\sqrt{k}\). Not possible unless p =1/3 , not allowed as this will make x not an integer ---> against the given statement. Keep. \(C)24\sqrt{3k}\). Possible with p=1. Eliminate. \(D)24\sqrt{6k}\). Possible with p=2. Eliminate. \(E)72\sqrt{k}\). Possible with p=3. Eliminate.

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