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:

1) says that x is divisible by 7 - but what about y ? It could be 0, or 1, each yielding a different result. We still don't know so statement 1 is insufficient.

2) xy is divisible by 49, so do we know anything about the respective values? No, xy can be 0 or 49 ... it's not suffcient

1+2) Aren't sufficient as well, since the case that y is 0 can't be excluded

This is one of the easy divisibility questions, will try to find a difficult one to explain the method

Don't know if the following question goes into the direction you want to, but give it try:

A certain number when successfully divided by 8 and 11 leaves remainders of 3 and 7 respectively. What will be remainder when the number is divided by the product of 8 and 11 ?

(1) 3
(2) 21
(3) 59
(4) 68

Formal answer as provided in the document available in this forum:

When a number is successfully divided by two divisors d1 and d2 and two remainders r1 and r2 are obtained, the remainder that will be obtained by the product of d1 and d2 is given by the relation

d1r2 + r1.

Where d1 and d2 are in ascending order respectively and r1 and r2 are their respective remainders when they divide the number.

In this case, the d1 = 8 and d2 = 11. And r1 = 3 and r2 = 7
Therefore, d1r2 + r1 = 8*7 + 3 = 59

Can you pls point me to the document that you mentioned.

allabout wrote:

Don't know if the following question goes into the direction you want to, but give it try:

A certain number when successfully divided by 8 and 11 leaves remainders of 3 and 7 respectively. What will be remainder when the number is divided by the product of 8 and 11 ?

(1) 3 (2) 21 (3) 59 (4) 68

Formal answer as provided in the document available in this forum:

When a number is successfully divided by two divisors d1 and d2 and two remainders r1 and r2 are obtained, the remainder that will be obtained by the product of d1 and d2 is given by the relation

d1r2 + r1.

Where d1 and d2 are in ascending order respectively and r1 and r2 are their respective remainders when they divide the number.

In this case, the d1 = 8 and d2 = 11. And r1 = 3 and r2 = 7 Therefore, d1r2 + r1 = 8*7 + 3 = 59

Hi, thanks for the reply... [#permalink]
26 Jan 2006, 14:10

This is something I havent seen before, thank you. Are there other formulas for figuring out if something is divisible by a certain number. I think my problem is that i don't know the best way to approach the questions.

allabout wrote:

At frist split up the given stem

xy*xy*y*y

1) says that x is divisible by 7 - but what about y ? It could be 0, or 1, each yielding a different result. We still don't know so statement 1 is insufficient.

2) xy is divisible by 49, so do we know anything about the respective values? No, xy can be 0 or 49 ... it's not suffcient

1+2) Aren't sufficient as well, since the case that y is 0 can't be excluded

This is one of the easy divisibility questions, will try to find a difficult one to explain the method

heres another example of a problem i have trouble with.... [#permalink]
26 Jan 2006, 21:10

just not sure the best way to approach these....any help is appreciated as I have a test coming up soon..
Thanks!
IF X is the integer, is X/3 an integer?

Re: whats the best way to approach divisibility questions? [#permalink]
26 Jan 2006, 21:55

ayl989 wrote:

I saw this question in an earlier post and always have problems with these.... Thanks! Is x^2*y^4 divisible by 49?

(1) x is an integer divisible by 7 (2) x*y is an integer divisible by 49.

Is x^2*y^4 divisible by 49?

(1) x is an integer divisible by 7
(2) x*y is an integer divisible by 49.

1) If x is an integer divisible by 7, it must be a multiple of 7. x can take values 7, 14, 21, 28... etc. Notice that x will be of the form 7*1, 7*2, 7*3.....7*n

So x^2 = 49n, x^2*y^4 = 49n(y^4). This will always be divisible by 49 since the numerator and denominator cancel out leaving you with just y^4. Y can be a fraction, an integer, a whole number, but it doesn't matter.

2) x*y = 49 -> This will result in (x,y) = (1,49), (49,1), (7,7) and no other combinations. All 3 combinations will be able to cancel out the 49.

thanks for the answers to the questions. but I was looking more for a method on how to approach these types of problems. what do you look to try for first? what order/method is used?

I just seem to get stuck and was wondering if anyone could provide pointers and not just find the answer.
Thanks again.

I'm not sure what you meant. But what I did previously was work a lot on these type of questions and then fine-tune my own method. There is no hard and fast way to solve such problems (probably the same thing with RC) but you just have to expose yourself to more of such questions and soon you'll feel much more comfortable facing them.