If x and y are positive integers, is xy a multiple of 8?

Question

If x and y are positive integers, is xy a multiple of 8 ?

(1) The greatest common divisor of x and y is 10.
(2) The least common multiple of x and y is 100.

Bunuel · Accepted Answer

(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.

Answer: C.

vitamingmat · Answer

It is good to see these kind of probs. thxs Bunuel ...

Yellow22 · Answer

Great explanation, thank you!

KarishmaB · Answer

An important property of LCM and GCF is 
x*y = GCD*LCM

(For explanation why this works, check out this link:  https://gmatclub.com/forum/gcf-lcm-ds-105745.html#p827452 )

Ques: Is xy divisible by 8?

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).

Sachin9 · Answer

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

KarishmaB · Answer

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.

nave · Answer

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. Hope this is clear

anamika7 · Answer

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.

Hence,   C  is the answer.

Aldorado · Answer

I believe these sort of questions can be handled without remembering the formula : X*Y = LCM (X, Y) * GCF (X,Y)

Let's see how,

Statement 1 : The greatest common divisor of x and y is 10

What this essentially means is that we have two numbers x and y, that can be written in the following way :
x = 2*5*A
y = 2*5*B , where A and B are co-primes. (If they had any common factor, it would've been considered in the GCF)
Now, x*y = 2*5*2*5*A*B => Now in order for x*y to be a factor of 8, we need at least one more 2 in either A or B; A condition can cannot be guaranteed just with the statement 1. Hence, insufficient.

Statement 2 : The least common multiple of x and y is 100

Statement 2 alone is not sufficient. Consider the below 2 examples :

Let x = 1 and y = 100. LCM (x,y) = 100 and x*y = 100 (not a multiple of 8)
Let x = 2 and y = 2*2*5*5. LCM (x,y) = 100 and x*y = 200 (multiple of 8)

Hence, statement 2 alone is not sufficient.

Now, considering statement 1 and 2 together :

Statement 1 : The greatest common divisor of x and y is 10
x = 2*5*A
y = 2*5*B , where A and B are co-primes.

Statement 2 : The least common multiple of x and y is 100
LCM (x,y) = 2*5 *A*B = 100
=> A*B = 10 
=> The only combination possible is 1*10 since A and B are co-primes.

Now , x*y = 2*5*2*5*2*5 (Always a multiple of 8).

Option C is the correct answer.
Would like to know your thoughts on using this approach while solving similar problems.

GMATinsight · Answer

Please check the solution as attached. Choosing Correct number and Property of LCM and HCF is all that is needed to answer the question

Answer: option C

BrentGMATPrepNow · Answer

Target question :  Is xy a multiple of 8?

Statement 1 : The greatest common factor of x and y is 10 
There are several values of x and y that satisfy statement 1. Here are two: 
case a: x=10 and y=10. Here, xy=100, which means  xy is NOT a multiple of 8 
case b: x=10 and y=20. Here, xy=200, which means  xy IS a multiple of 8 
Since we cannot answer the  target question  with certainty, statement 1 is NOT SUFFICIENT

Statement 2 : The least common multiple of x and y is 100 
There are several values of x and y that satisfy statement 2. Here are two: 
case a: x=1 and y=100. Here, xy=100, which means  xy s NOT a multiple of 8 
case b: x=2 and y=100. Here, xy=200, which means  xy IS a multiple of 8 
Since we cannot answer the  target question  with certainty, statement 2 is NOT SUFFICIENT

Statements 1 & 2  
There's a nice rule that says: 
  If x and y are positive integers, then (GCF of x and y)(LCM of x and y)=xy  
 (aside: whenever a question mentions the LCM and the GCF, be sure to consider the above rule) 
Statement 1 says the GCF of x and y is 10
Statement 2 says the LCM of x and y is 100
So, from our handy rule, xy = (10)(100) = 1,000, which means  xy IS a multiple of 8 
Since we can now answer the  target question  with certainty, the combined statements are SUFFICIENT

So, the answer is C
Cheers, 
Brent

Kritisood · Answer

Hi, i did as follows: (pls correct me if im wrong in my approach)
x*y= (2^3)k (k is any integer)

St 1: The greatest common factor of x and y is 10
10 =  2 *5 we need at least 2^3 for x*y to be divisible by 8 hence this statement may or may not be true

St 2: The least common multiple of x and y is 100
100 =  2^2 *5^2 again, this may or may not be true as we at least need 2^3 for x*y to be divisible by 8

Combined = x*y = LCM *HCF = 100*10 = 5^3* 2^3 
This is will always be true hence correct

Basshead · Answer

The question is asking if xy has  2^3  in its prime factorization.

From statement 1 we're told both x and y have 2*5 in its prime factorization. However, we don't know if either x or y has three 2's in its prime factorization. Statement 1 is insufficient.

Statement 2 tells us the LCM of x and y is  5^2 * 2^2 . From this statement we can conclude the highest power of 5 is 2 and the highest power of 2 is 2. But it's possible x has two 2's and y has a single 2. Statement 2 is insufficient.

Combined, we know the GCF of xy = 10. This means either x or y has two 2's and the other has a single 2. Therefore, we know for certain xy is a multiple of 8. SUFFICIENT.

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If x and y are positive integers, is xy a multiple of 8?

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