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Re: xy a multiple of 8 [#permalink]
09 Oct 2010, 11:01
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Orange08 wrote:
If x and y are positive integers, is xy a multiple of 8?
a) Greatest common divisor of x and y is 10 b) Least common factor of x and y is 100
(1) \(GCD(x,y)=10\) --> if \(x=10\) and \(y=10\), then \(xy=100\), which is not divisible by 8 BUT if \(x=10\) and \(y=20\), then \(xy=200\) which is divisible by 8. Two different answers. Not sufficient.
(2) \(LCM(x,y)=100\) --> if \(x=1\) and \(y=100\), then \(xy=100\), which is not divisible by 8 BUT if \(x=4\) and \(y=50\), then \(xy=200\) which is divisible by 8. Two different answers. Not sufficient.
(1)+(2) The most important property of LCM and GCD is: for any positive integers \(x\) and \(y\), \(x*y=GCD(x,y)*LCM(x,y)=10*100=1000\) --> 1000 is divisible by 8. Sufficient.
Stmnt 1: GCD = 10 If GCD is 10, LCM will be a multiple of 10 (the link above will explain why). Let us say LCM = 10a x*y = 10*10a = 100a We still cannot say whether xy is divisible by 8. Not sufficient.
Stmnt2: LCM = 100 If LCM is 100, we cannot say what GCD is. It could be 1 or 10 or 50 etc. x*y = GCD*100 We cannot say whether xy is divisible by 8. Not sufficient.
Taking both stmnts together, x*y = 10*100 = 1000 Since 1000 is divisible by 8, x*y is divisible by 8. Sufficient. Answer (C). _________________
Stmnt 1: GCD = 10 If GCD is 10, LCM will be a multiple of 10 (the link above will explain why). Let us say LCM = 10a x*y = 10*10a = 100a We still cannot say whether xy is divisible by 8. Not sufficient.
Stmnt2: LCM = 100 If LCM is 100, we cannot say what GCD is. It could be 1 or 10 or 50 etc. x*y = GCD*100 We cannot say whether xy is divisible by 8. Not sufficient.
Taking both stmnts together, x*y = 10*100 = 1000 Since 1000 is divisible by 8, x*y is divisible by 8. Sufficient. Answer (C).
Hi Karishma,
I cehcked the link but I did not understand why LCM will be a multiple of 10.. I understand that LCM*GCD=prod of 2 nos.. but don't understand why LCM has to be a multiple of 10..
Also, could you please help us with an approach on generating numbers for testing when a LCM is given.. in this case, lcm of x and y is 100.. .so how do we generate numbers whose lcm would be 100.
For gcd, we can write the GCD and multiply that by different numbers to get the pair of numbers.. Kindly explain how to do that for LCM _________________
hope is a good thing, maybe the best of things. And no good thing ever dies.
Re: xy a multiple of 8 [#permalink]
25 Jan 2013, 19:18
Expert's post
Sachin9 wrote:
Hi Karishma,
I cehcked the link but I did not understand why LCM will be a multiple of 10.. I understand that LCM*GCD=prod of 2 nos.. but don't understand why LCM has to be a multiple of 10..
Also, could you please help us with an approach on generating numbers for testing when a LCM is given.. in this case, lcm of x and y is 100.. .so how do we generate numbers whose lcm would be 100.
For gcd, we can write the GCD and multiply that by different numbers to get the pair of numbers.. Kindly explain how to do that for LCM
GIven that GCD = 10. What is GCD? It is the greatest common factor of two numbers i.e. both the numbers must have that factor. When you find the LCM of the numbers, the LCM includes all the factors of both the numbers. Hence, it will include 10 too. e.g. GCD = 10 Numbers: 10x, 10y where x and y are co-prime. What will be the LCM? LCM = 10xy (includes all factors of both the numbers)
In this question you don't need to list out the possible numbers given LCM = 100 but if you need to do it in another question, this is how you can handle that:
LCM = 100 = 2^2*5^2 Numbers: Split the primes -> (4, 25) Make one number = LCM -> (1, 100), (2, 100), (4, 100), (5, 100), (10, 100), (20, 100), (25, 100), (50, 100), (100, 100) One number must have the highest power of each prime -> (2^2*5, 2*5^2 which is 20, 50), (2^2, 2*5^2 which is 4, 50), (2^2*5, 5^2 which is 20, 25)
The overall strategy is this: Split the LCM into its prime factors. At least one number must have the highest power of each prime.
LCM = 2^a*3^b*5^c
At least one number must have 2^a, same or another number must have 3^b and same or another number must have 5^c. There are various possibilities. _________________
Stmnt 1: GCD = 10 If GCD is 10, LCM will be a multiple of 10 (the link above will explain why). Let us say LCM = 10a x*y = 10*10a = 100a We still cannot say whether xy is divisible by 8. Not sufficient.
Stmnt2: LCM = 100 If LCM is 100, we cannot say what GCD is. It could be 1 or 10 or 50 etc. x*y = GCD*100 We cannot say whether xy is divisible by 8. Not sufficient.
Taking both stmnts together, x*y = 10*100 = 1000 Since 1000 is divisible by 8, x*y is divisible by 8. Sufficient. Answer (C).
Hi Karishma,
I cehcked the link but I did not understand why LCM will be a multiple of 10.. I understand that LCM*GCD=prod of 2 nos.. but don't understand why LCM has to be a multiple of 10..
Also, could you please help us with an approach on generating numbers for testing when a LCM is given.. in this case, lcm of x and y is 100.. .so how do we generate numbers whose lcm would be 100.
For gcd, we can write the GCD and multiply that by different numbers to get the pair of numbers.. Kindly explain how to do that for LCM
I will look at this in a simpler manner.
GCF is the multiple of the lowest power of common factors (of x & y). LCM is the multiple of highest power of common factors. Therefore GCF is a factor is LCM. Since statement 1 says GCF is 10, so LCM can be assumed as multiple of 10.
Re: If x and y are positive integers, is xy a multiple of 8? (1) [#permalink]
06 Jul 2013, 19:54
For xy to be a multiple of 8 , xy should have minimum of 3 2's.
1. Gcd(x,y)=10 ===> x= 5*2 *m (where m can be any integer) y= 5*2* n (where n can be any integer) This does not clearly say that xy will be divisible by 8 because we are sure of only two 2's.
2. Lcm(x,y)= 100= 5^2* 2^2.====> It gives us different combinations of x and y. For eg: x= 5^2*2^2, y= 5*1, or x=5* 2^2, y= 5^2* 2 etc. Hence, xy may or may not be divisible by 8.
When 1+ 2 then, x= 5*2 , y= 5^2* 2^2 or vice versa. In any case we can see that xy is a multiple of 8.
Re: If x and y are positive integers, is xy a multiple of 8? (1) [#permalink]
17 Sep 2015, 07:11
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