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:

Official Answer and Stats are available only to registered users. Register/Login.

_________________

Encourage cooperation! If this post was very useful, kudos are welcome "It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

Re: If a and b are consecutive positive integers, and ab = 30x [#permalink]

Show Tags

01 Dec 2013, 13:45

1

This post received KUDOS

1

This post was BOOKMARKED

In order for ab = 30x = 2*3*5*x, ab needs to be divisible by 2,3 and 5 (because x can be any positive number from 1, 2...) -a and b are consecutive positive integer, so one must be even and the other is odd, so ab will be divisible by 2. -a^2 is divisible by 25 = 5^2, so a is divisible by 5. -63 = 3^2 * 7 is a factor of b^2, so b is divisible by 3 and 7. As you can see, the condition a alone is not sufficient because we only know ab will be divisible by 2 and 5. Condition b alone is not sufficient also because we only know ab will be divisible by 2, 3 and 7. If a and b together, we know ab = 2*5*3*7*something, so sufficient. Therefore, the correct answer is C.

If a and b are consecutive positive integers, and ab = 30x, is x an integer?

\(ab = 30x\) --> \(x=\frac{ab}{30}\). x to be an integer ab must be a multiple of 30.

(1) a^2 is divisible by 25. This implies that a is a multiple of 5. If a=5 and b=4, then ab=20 and x is NOT an integer but if a=5 and b=6, then ab=30 and x IS an integer. Not sufficient.

(2) 63 is a factor of b^2 --> \(63=3^2*7\). Thus b is a multiple of 3 (and 7). If b=21 and a=22, then ab=21*22 and x is NOT an integer but if b=21*30, then ab={a multiple of 30} and x IS and integer. Not sufficient.

(1)+(2) We have that a is a multiple of 5 and b is a multiple of 3. Also, since they are consecutive integers, then either one must be even. Therefore ab is a multiple of 2*3*5=30. Sufficient.

Re: If a and b are consecutive positive integers, and ab = 30x [#permalink]

Show Tags

19 Jun 2015, 08:21

I have one doubt. If b^2 has a factor 63 Then, b^2 has factors 3, 3, 7 for sure atleast once. Then, how can we be so sure that b has also 3 and 7 as factors. isnt b has only 3 and \sqrt{7} as factors?

Bunuel wrote:

If a and b are consecutive positive integers, and ab = 30x, is x an integer?

\(ab = 30x\) --> \(x=\frac{ab}{30}\). x to be an integer ab must be a multiple of 30.

(1) a^2 is divisible by 25. This implies that a is a multiple of 5. If a=5 and b=4, then ab=20 and x is NOT an integer but if a=5 and b=6, then ab=30 and x IS an integer. Not sufficient.

(2) 63 is a factor of b^2 --> \(63=3^2*7\). Thus b is a multiple of 3 (and 7). If b=21 and a=22, then ab=21*22 and x is NOT an integer but if b=21*30, then ab={a multiple of 30} and x IS and integer. Not sufficient.

(1)+(2) We have that a is a multiple of 5 and b is a multiple of 3. Also, since they are consecutive integers, then either one must be even. Therefore ab is a multiple of 2*3*5=30. Sufficient.

I have one doubt. If b^2 has a factor 63 Then, b^2 has factors 3, 3, 7 for sure atleast once. Then, how can we be so sure that b has also 3 and 7 as factors. isnt b has only 3 and \sqrt{7} as factors?

Bunuel wrote:

If a and b are consecutive positive integers, and ab = 30x, is x an integer?

\(ab = 30x\) --> \(x=\frac{ab}{30}\). x to be an integer ab must be a multiple of 30.

(1) a^2 is divisible by 25. This implies that a is a multiple of 5. If a=5 and b=4, then ab=20 and x is NOT an integer but if a=5 and b=6, then ab=30 and x IS an integer. Not sufficient.

(2) 63 is a factor of b^2 --> \(63=3^2*7\). Thus b is a multiple of 3 (and 7). If b=21 and a=22, then ab=21*22 and x is NOT an integer but if b=21*30, then ab={a multiple of 30} and x IS and integer. Not sufficient.

(1)+(2) We have that a is a multiple of 5 and b is a multiple of 3. Also, since they are consecutive integers, then either one must be even. Therefore ab is a multiple of 2*3*5=30. Sufficient.

Answer: C.

Hope it's clear.

Hi, b is a positive integer so it cannot be3*\(\sqrt{7}\)* something else... therefore it has to be a multiple of 7 and not \(\sqrt{7}\)
_________________

I have one doubt. If b^2 has a factor 63 Then, b^2 has factors 3, 3, 7 for sure atleast once. Then, how can we be so sure that b has also 3 and 7 as factors. isnt b has only 3 and \sqrt{7} as factors?

Bunuel wrote:

If a and b are consecutive positive integers, and ab = 30x, is x an integer?

\(ab = 30x\) --> \(x=\frac{ab}{30}\). x to be an integer ab must be a multiple of 30.

(1) a^2 is divisible by 25. This implies that a is a multiple of 5. If a=5 and b=4, then ab=20 and x is NOT an integer but if a=5 and b=6, then ab=30 and x IS an integer. Not sufficient.

(2) 63 is a factor of b^2 --> \(63=3^2*7\). Thus b is a multiple of 3 (and 7). If b=21 and a=22, then ab=21*22 and x is NOT an integer but if b=21*30, then ab={a multiple of 30} and x IS and integer. Not sufficient.

(1)+(2) We have that a is a multiple of 5 and b is a multiple of 3. Also, since they are consecutive integers, then either one must be even. Therefore ab is a multiple of 2*3*5=30. Sufficient.

Answer: C.

Hope it's clear.

1. Only positive integers are considered as factors. So, saying that \(\sqrt{7}\) is a factor of some integer does not make sense.

2. Exponentiation does not "produce" primes, meaning that for positive integers m and n, m^n will have the same exact prime factors as m itself does. Thus, if 3 and 7 are factors of b^2 it must be true that both of them are factors of b too (else how would they appear in b^2).

Re: If a and b are consecutive positive integers, and ab = 30x [#permalink]

Show Tags

26 Sep 2017, 17:16

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________