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:

Hi
Stmnt 1 is not SUFF
N would be an integer that ends with 9. But a n integer is divisible by 4 if last 2 digits are divisible by 4. If N is 9 rem is 1, If N is 19 rem is 3 so Stmnt 1 is NOT suff

St1: Clearly INSUFF
St2:
If p-q = 10
pq-(p+q) can be written as q^2+8q-10
or (q^2 - 1) + (8q-9)
8q-9 will give a remainder of 3 when divided by 4
now lets see q^2 - 1
q^2 - 1 = (q-1) (q+1)
since q is odd both (q-1) and (q+1) are divisible by 2 hence it is divisible by 4.
Final remainder = 3: SUFF

St1: Clearly INSUFF St2: If p-q = 10 pq-(p+q) can be written as q^2+8q-10 or (q^2 - 1) + (8q-9) 8q-9 will give a remainder of 3 when divided by 4 now lets see q^2 - 1 q^2 - 1 = (q-1) (q+1) since q is odd both (q-1) and (q+1) are divisible by 2 hence it is divisible by 4. Final remainder = 3: SUFF

Why is (1) clearly insufficient? Shouldn't you know me better by now?

St1: Clearly INSUFF St2: If p-q = 10 pq-(p+q) can be written as q^2+8q-10 or (q^2 - 1) + (8q-9) 8q-9 will give a remainder of 3 when divided by 4 now lets see q^2 - 1 q^2 - 1 = (q-1) (q+1) since q is odd both (q-1) and (q+1) are divisible by 2 hence it is divisible by 4. Final remainder = 3: SUFF

Why is (1) clearly insufficient? Shouldn't you know me better by now?

Oops...
Answer should be D.
St1:
p = 10x +1
q = 10y+1
where x and y are positive integers

Now pq-(p+q) can be written as
(10x+1) (10y+1) - 10x-1-10y-1
100xy +10x+10y +1 - 10x-1-10y-1
100xy - 1 so the remainder will always be 3: SUFF

If something is divisible by 4 and then you subtract 1 from that then we need to subtract 3 more from that to again make it divisible by 4. Hence remainder = 3

Same way if something is divisible by 4 and then you subtract 9 from that then we need to subtract 3 more from that to again make it divisible by 4. Hence remainder = 3