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If x and y are positive integers, what is the value of xy? [#permalink]
22 Jun 2008, 11:05

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If x and y are positive integers, what is the value of xy? 1) The greatest common factor of x and y is 10 2) the least common multiple of x and y is 180

Re: DS - GMATPrep - value of xy [#permalink]
25 Jun 2008, 11:29

1

This post received KUDOS

In the solution above, part 2) is a bit incomplete. Yes, you do not know whether x or y (or both) is divisible by 5, but that's not the only issue: you also do not know whether each is divisible by 2^2 or by 3^2.

You certainly know that one of x or y is divisible by 2^2. The other might be divisible either by 2^0, 2^1 or 2^2. You certainly know that one of x or y is divisible by 3^2. The other might be divisible either by 3^0, 3^1 or 3^2 This leads to many possibilities, and many possible values for the product xy.

From Statement 2) alone, we can say is that the product xy is greater than or equal to 180, and less than or equal to 180^2, but there are many possible values for xy. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Re: DS - GMATPrep - value of xy [#permalink]
25 Jun 2008, 06:41

x-ALI-x wrote:

Thanks greenoak, I didn't know this rule. Is there another way to solve this problem?

There are other ways to look at this problem, but I'm not sure why one would want to use a different method. The rule posted by greenoak is a fundamental law of number theory, and it lets you answer this question (and other GMAT questions) in a few seconds. It's definitly worth understanding why the rule greenoak posted is true- if you understand LCMs and GCDs, the rule should make sense.

Alternatively, you might choose numbers, as maximusdecimus suggests. There aren't many possibilities to consider, since we only need to look at multiples of 10 which are also divisors of 180. We know one of the divisors must be divisible by 9, while the other cannot be divisible by 3 -- otherwise the GCD would be 30 or 90. The only possible values for x and y are 10 and 180; or 20 and 90 (in either order). The product is 1800 in each case.

Or, you could, from statement 1, conclude that x/10 and y/10 are integers which share no divisors- they are 'relatively prime' in math speak. Thus, the LCM of x/10 and y/10 is equal to the product of x/10 and y/10. And from statement 2, the LCM of x/10 and y/10 must be 18. Thus (x/10)*(y/10) = 18 and xy = 1800. But all I've done here is prove a special case of the general rule that greenoak posted, and there's more value in understanding the general rule than the specific case. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Re: DS - GMATPrep - value of xy [#permalink]
25 Jun 2008, 12:31

IanStewart wrote:

In the solution above, part 2) is a bit incomplete. Yes, you do not know whether x or y (or both) is divisible by 5, but that's not the only issue: you also do not know whether each is divisible by 2^2 or by 3^2.

You certainly know that one of x or y is divisible by 2^2. The other might be divisible either by 2^0, 2^1 or 2^2. You certainly know that one of x or y is divisible by 3^2. The other might be divisible either by 3^0, 3^1 or 3^2 This leads to many possibilities, and many possible values for the product xy.

From Statement 2) alone, we can say is that the product xy is greater than or equal to 180, and less than or equal to 180^2, but there are many possible values for xy.

Yes, good point. I didn't even think of that. +1

I have edited my above response by incorporating your logic into my rationale.

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Factorials were someone's attempt to make math look exciting!!!

Re: DS - GMATPrep - value of xy [#permalink]
26 Jun 2008, 14:55

x-ALI-x wrote:

If x and y are positive integers, what is the value of xy? 1) The greatest common factor of x and y is 10 2) the least common multiple of x and y is 180

The OA is C. Please explain!!!!!

This troubled me for sometime. Here is another way to solve the problem.

If we consider two numbers, lets say 18 and 24, the LCM (denoted by L) is calculated as follows.

L = 18a ; L = 24b (a and b are integers)

L = 2*3*3*a ; L = 2*2*2*3*b

so, taking the bold factors out from the above (the product of bold factors constitute GCF, denoted by G); we can say that :

a = 2*2 b= 3

rewriting :

L = G*3*a L= G*4*b

using the same logic to the given problem :

We all agree that both statements are insufficient to answer the question. Considering both together :

Re: DS - GMATPrep - value of xy [#permalink]
26 Jun 2008, 21:35

sanjay_gmat wrote:

x-ALI-x wrote:

If x and y are positive integers, what is the value of xy? 1) The greatest common factor of x and y is 10 2) the least common multiple of x and y is 180

The OA is C. Please explain!!!!!

This troubled me for sometime. Here is another way to solve the problem.

If we consider two numbers, lets say 18 and 24, the LCM (denoted by L) is calculated as follows.

L = 18a ; L = 24b (a and b are integers)

L = 2*3*3*a ; L = 2*2*2*3*b

so, taking the bold factors out from the above (the product of bold factors constitute GCF, denoted by G); we can say that :

a = 2*2 b= 3

rewriting :

L = G*3*a L= G*4*b

using the same logic to the given problem :

We all agree that both statements are insufficient to answer the question. Considering both together :

L = G*f2*f1 (x = G*f2) L = G*f1*f2 (y = G*f1)

180 = 10*f1*f2

=> f1*f2 = 18

since x = G*f2 and y = G*f1 , we have :

x*y = G^2 * f1*f2

= 10^2 * 18 = 1800

Hence C.

Also, from the last equation :

x*y = G*G*f1*f2 = G*L

this proves the equation that greenoak used.

gmatclubot

Re: DS - GMATPrep - value of xy
[#permalink]
26 Jun 2008, 21:35