Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Using (1), x is also a multiple of 9. Then smallest possible value of X is 18. We can't tell whether xy is a multiple of 105 as we cannot determine if there is a prime factor 5 present in Y.

Using (2), y is a multple of 25. So we have prime factors 7, 5, 2, and 3. So xy is a multiple of 105.

x is a multiple of 6==> x has at least two factors, 2 & 3 y is a multiple of14==> y has at least two factors, 2 & 7

xy will be a multiple of 105, if the factors of xy (combined factors of x & y) can be formed together to make 105.

Statement 1: x is a multiple of 9==> In addtion to 2 & 3 , x also has one more factor 3. So x has at least three factors 2,3 & 3. Now if we multiply x & y and hence multiply the possible factors of x & y we will have the following possible factors of xy: 2,3,3,2 & 7. Now we can't form 105 by multiplying any of these numbers. Hence we can say that statement 1 is not sufficient.

Statement 2: y is a multiple of 25>> In addition to 2 & 7, y also has two more factors, 5 & 5 or 5^2. Now if we multiply x & y and hence multiply the possible factors of x & y we will have the following possible group 2,3,7,5 & 5. You can see that we can easily form 105 by multiplying 3 , 7 & 5. Statement 2 is sufficient

Hence answer is "B".

This is my way of thinking. Kindly correct me if I am wrong. _________________

Although the answer is correct. The method is wrong

AKProdigy87 wrote:

a) x = 6n = (3)(2)n b) y = 14n = (7)(2)n

If x= 6n, you cannot assume that y= 14n; y could also be 14m

AKProdigy87 wrote:

We know that therefore: xy = (3)(7)(2)(2)n

xy need not be a equal to (3)(7)(2)(2)n; xy is a multiple of (3)(7)(2) =42

I used n to represent an integer (any integer)... not as a means of equating that the integer n was the same in both cases. I can see how the confusion could arise though.

If positive integer x is a multiple of 6, and positive integer y is a multiple of 14, is xy a multiple of 105? x = 2 * 3 * a y = 2 * 7 * b xy = 3 * 5 * 7 * c???? where a,b,c etc are integers (1) x is a multiple of 9 x = 3*3 * d .. so from the above equations x can be written as x=2*3*3 * e the product of xy can be written as xy = 2*3*3*7 * f if f is a multiple of 5 then xy will be multiple of 105 else no Insufficient (2) y is a multiple of 25 y = 5*5*g ... so from the above equations y can be written as y=2*7*5*5 *h the product of xy can be written as xy = 2*3*5*5*7 * i = 3*5*7 * 2*5*i = 105 * 2*5*i xy is a multiple of 105 Sufficient

It seems I'm quite weak with multiples and factors, even after reading the explanation I still don't get it

If positive integer x is a multiple of 6, and positive integer y is a multiple of 14, is xy a multiple of 105? (1) x is a multiple of 9 (2) y is a multiple of 25

It seems I'm quite weak with multiples and factors, even after reading the explanation I still don't get it

If positive integer x is a multiple of 6, and positive integer y is a multiple of 14, is xy a multiple of 105? (1) x is a multiple of 9 (2) y is a multiple of 25

if x is multiple of 6 and y is multiple of 14 xy is definitely multiple of 2*3*7 (as this is the LCM of 6 and 14) now 105 comprises 5*3*7 so xy has to have 5*3*7 as its factor in order to be the multiple of 105 stmt1 says x is a multiple of 9 dat means xy is multiple of 2*3*3*7 for sure but this does not prove if it is multiple of 105 or not stmt2 says y is a multiple of 25 so xy will definitely be multiple of 2*3*7*5*5 and this LCM contains 105 in it which proves that xy is a multiple of 105.

if X is a multiple of 6 and 9 both as per statment 1, then X should be 18, 36, 72 and so on....... in any case xy is not a multiple of 105. Hence, sufficient

if X is a multiple of 6 and 9 both as per statment 1, then X should be 18, 36, 72 and so on....... in any case xy is not a multiple of 105. Hence, sufficient

Am i doing anything wrong here?

X is a multiple of 6 and y is a multiple of 14. but X or y can be multiple of other numbers also like 4,5,11 etc. xy is definitely a multiple of 2*3*7 = 42 (Taking the LCM) but if we need to find if xy is a multiple of 18 we have to be sure that xy contains 5*3*7 at least. Since xy does contain 3,7, so it can be multiple of 105 only if it also has 5 as a multiple too. Only second statement states that y is also a multiple of 25. which means XY will be multiple of LCM(6, 14, 25) = 2*3*7*5*5 = 1050 since XY is multiple of 1050 its definitely a multiple of 105. _________________

If positive integer x is a multiple of 6 and positive integer y is a multiple of 14, is xy a multiple of 105 ? (1) x is a multiple of 9. (2) y is a multiple of 25.

Can someone explain this

\(105=3*5*7\). Since \(x\) is a multiple of 6 and \(y\) is a multiple of 14, then \(xy\) is a multiple of \(LCM(x,y)=2*3*7\): we have 3 and 7 as factors of \(xy\), so in order \(xy\) to be a multiple of 105 we need missing 5 to be a factor of either \(x\) or \(y\).

(1) x is a multiple of 9 --> we don't know whether 5 is a factor of either \(x\) or \(y\). Not sufficient.

(2) y is a multiple of 25 --> 5 is a factor of \(y\), hence \(xy\) is a multiple of 105. Sufficient.

Final decisions are in: Berkeley: Denied with interview Tepper: Waitlisted with interview Rotman: Admitted with scholarship (withdrawn) Random French School: Admitted to MSc in Management with scholarship (...

Last year when I attended a session of Chicago’s Booth Live , I felt pretty out of place. I was surrounded by professionals from all over the world from major...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...