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If x and y are positive integers, what is the value of xy? [#permalink]
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22 Jun 2008, 12:05
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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
The OA is C. Please explain!!!!!



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Re: DS  GMATPrep  value of xy [#permalink]
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22 Jun 2008, 13:06
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Each statement by itself is clearly insufficient.
For both:
x*y = LCM(x,y)*GCF(x,y) (always true for positive numbers). Thus, C.



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Re: DS  GMATPrep  value of xy [#permalink]
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25 Jun 2008, 12:29
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.
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Re: DS  GMATPrep  value of xy [#permalink]
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22 Jun 2008, 15:17
Thanks greenoak, I didn't know this rule. Is there another way to solve this problem?



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Re: DS  GMATPrep  value of xy [#permalink]
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24 Jun 2008, 17:54
Does anyone have another workaround for this question other than the rule posted by greenoak?



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Re: DS  GMATPrep  value of xy [#permalink]
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25 Jun 2008, 05:15
I am not sure but another way may be like: 1)gives these numbers are like: (20,30),(10,90),(30,70),... then 1) & 2) give (20,90). chango wrote: Does anyone have another workaround for this question other than the rule posted by greenoak?



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Re: DS  GMATPrep  value of xy [#permalink]
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25 Jun 2008, 07:41
xALIx 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.
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Re: DS  GMATPrep  value of xy [#permalink]
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25 Jun 2008, 10:32
chango wrote: Does anyone have another workaround for this question other than the rule posted by greenoak? Not to take away what greenoak posted, because that is the best way to solve this problem, but if you want another perspective on it, see below:
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Re: DS  GMATPrep  value of xy [#permalink]
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25 Jun 2008, 13:31
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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Re: DS  GMATPrep  value of xy [#permalink]
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26 Jun 2008, 15:55
xALIx 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.



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Re: DS  GMATPrep  value of xy [#permalink]
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26 Jun 2008, 22:35
sanjay_gmat wrote: xALIx 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.




Re: DS  GMATPrep  value of xy
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