It is currently 21 Nov 2017, 10:42

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If p and q are two different prime numbers and n is the

Author Message
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1259

Kudos [?]: 341 [0], given: 0

If p and q are two different prime numbers and n is the [#permalink]

### Show Tags

12 Apr 2007, 07:18
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If p and q are two different prime numbers and n is the difference between their product and their sum, what is the remainder when n is divided by 4?

(1) Both p and q have a units digit of 1.
(2) p-q= 10

Kudos [?]: 341 [0], given: 0

Intern
Joined: 29 Aug 2005
Posts: 4

Kudos [?]: [0], given: 0

### Show Tags

12 Apr 2007, 07:38
Statement (1) Both p and q have a units digit of 1.
31, 41, 61, 71 ... So not sufficient

Statement (2) p-q= 10
23-13 = 10
41-31 = 10
71-61 = 10
So not sufficient

edit...
Statements 1 and 2 together
n=31*41-(31+41) = 1199
or
n=61*71-(61+71) = 4199

Reminder is 3 when n is devided by 4 for both the values of n.

So Answer should be C. (Please correct me if I am wrong.)

Last edited by newgmater on 12 Apr 2007, 07:47, edited 2 times in total.

Kudos [?]: [0], given: 0

Director
Joined: 13 Nov 2003
Posts: 788

Kudos [?]: 64 [0], given: 0

Location: BULGARIA

### Show Tags

12 Apr 2007, 07:40
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

sTMNT 2 is SUFF

ANS B

Kudos [?]: 64 [0], given: 0

CEO
Joined: 20 Nov 2005
Posts: 2892

Kudos [?]: 336 [0], given: 0

Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008

### Show Tags

12 Apr 2007, 08:13
B

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

Kudos [?]: 336 [0], given: 0

Intern
Joined: 29 Aug 2005
Posts: 4

Kudos [?]: [0], given: 0

### Show Tags

12 Apr 2007, 08:38
ps_dahiya. I am missing something. Could you please explain how you are writing pq - (p+q) = q^2+8q-10 (provided p-q=10)

Thank you.

Kudos [?]: [0], given: 0

CEO
Joined: 20 Nov 2005
Posts: 2892

Kudos [?]: 336 [0], given: 0

Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008

### Show Tags

12 Apr 2007, 08:55
newgmater wrote:
ps_dahiya. I am missing something. Could you please explain how you are writing pq - (p+q) = q^2+8q-10 (provided p-q=10)

Thank you.

n = pq-(p+q)
substitute p= q+10 in above equation and you get q^2+8q-10

hope this helps...

Kudos [?]: 336 [0], given: 0

GMAT Instructor
Joined: 04 Jul 2006
Posts: 1259

Kudos [?]: 341 [0], given: 0

### Show Tags

12 Apr 2007, 10:30
ps_dahiya wrote:
B

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?

Kudos [?]: 341 [0], given: 0

CEO
Joined: 20 Nov 2005
Posts: 2892

Kudos [?]: 336 [0], given: 0

Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008

### Show Tags

12 Apr 2007, 11:27
kevincan wrote:
ps_dahiya wrote:
B

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...
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

A little out of practice...

Kudos [?]: 336 [0], given: 0

GMAT Instructor
Joined: 04 Jul 2006
Posts: 1259

Kudos [?]: 341 [0], given: 0

### Show Tags

13 Apr 2007, 02:56
Very nice! Also note that for prime numbers p and q, n=pq-p-q=(p-1)(q-1)-1

Kudos [?]: 341 [0], given: 0

Intern
Joined: 10 Mar 2007
Posts: 4

Kudos [?]: [0], given: 0

### Show Tags

13 Apr 2007, 17:41
My first post.

Sorry, how did you get the reminder 3?

Thanx

Kudos [?]: [0], given: 0

CEO
Joined: 20 Nov 2005
Posts: 2892

Kudos [?]: 336 [0], given: 0

Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008

### Show Tags

13 Apr 2007, 22:26
jaspetrovic wrote:
My first post.

Sorry, how did you get the reminder 3?

Thanx

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

Kudos [?]: 336 [0], given: 0

13 Apr 2007, 22:26
Display posts from previous: Sort by