GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Nov 2018, 20:39

### 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 November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### FREE Quant Workshop by e-GMAT!

November 18, 2018

November 18, 2018

07:00 AM PST

09:00 AM PST

Get personalized insights on how to achieve your Target Quant Score. November 18th, 7 AM PST
• ### How to QUICKLY Solve GMAT Questions - GMAT Club Chat

November 20, 2018

November 20, 2018

09:00 AM PST

10:00 AM PST

The reward for signing up with the registration form and attending the chat is: 6 free examPAL quizzes to practice your new skills after the chat.

# If p and n are positive integers and p > n, what is the

Author Message
TAGS:

### Hide Tags

Intern
Joined: 20 Feb 2012
Posts: 32
If p and n are positive integers and p > n, what is the  [#permalink]

### Show Tags

23 Feb 2012, 07:10
9
66
00:00

Difficulty:

85% (hard)

Question Stats:

55% (01:20) correct 45% (01:12) wrong based on 1248 sessions

### HideShow timer Statistics

If p and n are positive integers and p > n, what is the remainder when p^2 - n^2 is divided by 15 ?

(1) The remainder when p + n is divided by 5 is 1.
(2) The remainder when p - n is divided by 3 is 1.
Math Expert
Joined: 02 Sep 2009
Posts: 50623

### Show Tags

23 Feb 2012, 07:20
34
50
If p and n are positive integers and p>n, what is the remainder when p^2 - n^2 is divided by 15?

First of all $$p^2 - n^2=(p+n)(p-n)$$.

(1) The remainder when p + n is divided by 5 is 1. No info about p-n. Not sufficient.

(2) The remainder when p - n is divided by 3 is 1. No info about p+n. Not sufficient.

(1)+(2) "The remainder when p + n is divided by 5 is 1" can be expressed as $$p+n=5t+1$$ and "The remainder when p - n is divided by 3 is 1" can be expressed as $$p-n=3k+1$$.

Multiply these two --> $$(p+n)(p-n)=(5t+1)(3k+1)=15kt+5t+3k+1$$, now first term (15kt) is clearly divisible by 15 (r=0), but we don't know about 5t+3k+1. For example t=1 and k=1, answer r=9 BUT t=7 and k=3, answer r=0. Not sufficient.

OR by number plugging: if $$p+n=11$$ (11 divided by 5 yields remainder of 1) and $$p-n=1$$ (1 divided by 3 yields remainder of 1) then $$(p+n)(p-n)=11$$ and remainder upon division 11 by 15 is 11 BUT if $$p+n=21$$ (21 divided by 5 yields remainder of 1) and $$p-n=1$$ (1 divided by 3 yields remainder of 1) then $$(p+n)(p-n)=21$$ and remainder upon division 21 by 15 is 6. Not sufficient.

_________________
##### General Discussion
Manager
Joined: 22 Dec 2011
Posts: 238

### Show Tags

04 Nov 2012, 21:16
Bunuel wrote:
If p and n are positive integers and p>n, what is the remainder when p^2 - n^2 is divided by 15?

First of all $$p^2 - n^2=(p+n)(p-n)$$.

(1) The remainder when p + n is divided by 5 is 1. No info about p-n. Not sufficient.

(2) The remainder when p - n is divided by 3 is 1. No info about p+n. Not sufficient.

(1)+(2) "The remainder when p + n is divided by 5 is 1" can be expressed as $$p+n=5t+1$$ and "The remainder when p - n is divided by 3 is 1" can be expressed as $$p-n=3k+1$$.

Multiply these two --> $$(p+n)(p-n)=(5t+1)(3k+1)=15kt+5t+3k+1$$, now first term (15kt) is clearly divisible by 15 (r=0), but we don't know about 5t+3k+1. For example t=1 and k=1, answer r=9 BUT t=7 and k=3, answer r=0. Not sufficient.

OR by number plugging: if $$p+n=11$$ (11 divided by 5 yields remainder of 1) and $$p-n=1$$ (1 divided by 3 yields remainder of 1) then $$(p+n)(p-n)=11$$ and remainder upon division 11 by 15 is 11 BUT if $$p+n=21$$ (21 divided by 5 yields remainder of 1) and $$p-n=1$$ (1 divided by 3 yields remainder of 1) then $$(p+n)(p-n)=21$$ and remainder upon division 21 by 15 is 6. Not sufficient.

Hi Bunuel - 1 doubt.. why can't the below process be followed?

p+n = 5A+1 => 1,6,11,16,21,26
p-n = 3B+1 => 1,4,7,10,13,16,19,21

p+n * p-n => 15 K + 16. Hence the remainder on division by 15 gives 1.

Cheers
Manager
Joined: 10 Jan 2011
Posts: 158
Location: India
GMAT Date: 07-16-2012
GPA: 3.4
WE: Consulting (Consulting)

### Show Tags

05 Nov 2012, 03:33
1
Jp27 wrote:
Bunuel wrote:
If p and n are positive integers and p>n, what is the remainder when p^2 - n^2 is divided by 15?

First of all $$p^2 - n^2=(p+n)(p-n)$$.

(1) The remainder when p + n is divided by 5 is 1. No info about p-n. Not sufficient.

(2) The remainder when p - n is divided by 3 is 1. No info about p+n. Not sufficient.

(1)+(2) "The remainder when p + n is divided by 5 is 1" can be expressed as $$p+n=5t+1$$ and "The remainder when p - n is divided by 3 is 1" can be expressed as $$p-n=3k+1$$.

Multiply these two --> $$(p+n)(p-n)=(5t+1)(3k+1)=15kt+5t+3k+1$$, now first term (15kt) is clearly divisible by 15 (r=0), but we don't know about 5t+3k+1. For example t=1 and k=1, answer r=9 BUT t=7 and k=3, answer r=0. Not sufficient.

OR by number plugging: if $$p+n=11$$ (11 divided by 5 yields remainder of 1) and $$p-n=1$$ (1 divided by 3 yields remainder of 1) then $$(p+n)(p-n)=11$$ and remainder upon division 11 by 15 is 11 BUT if $$p+n=21$$ (21 divided by 5 yields remainder of 1) and $$p-n=1$$ (1 divided by 3 yields remainder of 1) then $$(p+n)(p-n)=21$$ and remainder upon division 21 by 15 is 6. Not sufficient.

Hi Bunuel - 1 doubt.. why can't the below process be followed?

p+n = 5A+1 => 1,6,11,16,21,26
p-n = 3B+1 => 1,4,7,10,13,16,19,21

p+n * p-n => 15 K + 16. Hence the remainder on division by 15 gives 1.

Cheers

try picking numbers:

if p= 5 and n = 1 p+n = 6 remainder after dividing by 5 is 1
p-n= 4 remainder after dividing by 3 is 1

p^2 -N2 = 25-1 = 24 remainder after dividing by 15 is 9

now consider p =9 and n =2

p+n = 11 remainder after dividing by 5 is 1
p-n = 7 remainder after dividing by 3 is 1

p^2 - N^2 = 81-4 = 77 remainder after dividing by 15 is 2.

hence both statement combined are not suff
_________________

-------Analyze why option A in SC wrong-------

Math Expert
Joined: 02 Sep 2009
Posts: 50623

### Show Tags

06 Nov 2012, 04:21
1
1
Jp27 wrote:
Bunuel wrote:
If p and n are positive integers and p>n, what is the remainder when p^2 - n^2 is divided by 15?

First of all $$p^2 - n^2=(p+n)(p-n)$$.

(1) The remainder when p + n is divided by 5 is 1. No info about p-n. Not sufficient.

(2) The remainder when p - n is divided by 3 is 1. No info about p+n. Not sufficient.

(1)+(2) "The remainder when p + n is divided by 5 is 1" can be expressed as $$p+n=5t+1$$ and "The remainder when p - n is divided by 3 is 1" can be expressed as $$p-n=3k+1$$.

Multiply these two --> $$(p+n)(p-n)=(5t+1)(3k+1)=15kt+5t+3k+1$$, now first term (15kt) is clearly divisible by 15 (r=0), but we don't know about 5t+3k+1. For example t=1 and k=1, answer r=9 BUT t=7 and k=3, answer r=0. Not sufficient.

OR by number plugging: if $$p+n=11$$ (11 divided by 5 yields remainder of 1) and $$p-n=1$$ (1 divided by 3 yields remainder of 1) then $$(p+n)(p-n)=11$$ and remainder upon division 11 by 15 is 11 BUT if $$p+n=21$$ (21 divided by 5 yields remainder of 1) and $$p-n=1$$ (1 divided by 3 yields remainder of 1) then $$(p+n)(p-n)=21$$ and remainder upon division 21 by 15 is 6. Not sufficient.

Hi Bunuel - 1 doubt.. why can't the below process be followed?

p+n = 5A+1 => 1,6,11,16,21,26
p-n = 3B+1 => 1,4,7,10,13,16,19,21

p+n * p-n => 15 K + 16. Hence the remainder on division by 15 gives 1.

Cheers

p+n = 5A+1 and p-n = 3B+1 does not mean that (p+n)*(p-n)=15K+16. When you expand (p+n)*(p-n)=(5A+1)(3B+1) you won't get an expression of the form 15K+16.
_________________
Intern
Joined: 23 Oct 2012
Posts: 28

### Show Tags

13 Dec 2013, 01:34
Bunuel wrote:
If p and n are positive integers and p>n, what is the remainder when p^2 - n^2 is divided by 15?

First of all $$p^2 - n^2=(p+n)(p-n)$$.

(1) The remainder when p + n is divided by 5 is 1. No info about p-n. Not sufficient.

(2) The remainder when p - n is divided by 3 is 1. No info about p+n. Not sufficient.

(1)+(2) "The remainder when p + n is divided by 5 is 1" can be expressed as $$p+n=5t+1$$ and "The remainder when p - n is divided by 3 is 1" can be expressed as $$p-n=3k+1$$.

Multiply these two --> $$(p+n)(p-n)=(5t+1)(3k+1)=15kt+5t+3k+1$$, now first term (15kt) is clearly divisible by 15 (r=0), but we don't know about 5t+3k+1. For example t=1 and k=1, answer r=9 BUT t=7 and k=3, answer r=0. Not sufficient.

OR by number plugging: if $$p+n=11$$ (11 divided by 5 yields remainder of 1) and $$p-n=1$$ (1 divided by 3 yields remainder of 1) then $$(p+n)(p-n)=11$$ and remainder upon division 11 by 15 is 11 BUT if $$p+n=21$$ (21 divided by 5 yields remainder of 1) and $$p-n=1$$ (1 divided by 3 yields remainder of 1) then $$(p+n)(p-n)=21$$ and remainder upon division 21 by 15 is 6. Not sufficient.

Hi Bunuel,

I reasoned this question as follows:

Prompt is asking what is remainder when... (p+n) (p-n) /15 ?

St 1> gives remainder when p+n is divided by 5 is 1...this implies..remainder when (p+n) is divided by 15 is also 1...(since 5 is a factor of 15). ..NOT SUFF
St 2> gives remainder when p-n is divided by 3 is 1...this implies..remainder when (p-n) is divided by 15 is also 1...(since 3 is a factor of 15) ..not SUFF

Combinign 1 & 2... (p+n) /15 gives remainder 1, and (p-n) /15 gives remainder 1.....so (p+n) (p-n)/15 ...should also yield remainder 1...since we can multiply the remainders in this case...because the divisor is the same. . Please clarify, is this correct?? If not , why?
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8550
Location: Pune, India

### Show Tags

14 Dec 2013, 00:03
audiogal101 wrote:

Prompt is asking what is remainder when... (p+n) (p-n) /15 ?

St 1> gives remainder when p+n is divided by 5 is 1...this implies..remainder when (p+n) is divided by 15 is also 1...(since 5 is a factor of 15). ..NOT SUFF
St 2> gives remainder when p-n is divided by 3 is 1...this implies..remainder when (p-n) is divided by 15 is also 1...(since 3 is a factor of 15) ..not SUFF

Combinign 1 & 2... (p+n) /15 gives remainder 1, and (p-n) /15 gives remainder 1.....so (p+n) (p-n)/15 ...should also yield remainder 1...since we can multiply the remainders in this case...because the divisor is the same. . Please clarify, is this correct?? If not , why?

Responding to a pm:

The highlighted portion is incorrect.

Say p + n = 21.
When it is divided by 5, the remainder is 1. But when it is divided by 15, the remainder is 6. So your implication is not correct. You cannot say that the remainder when you divide p+n by 15 will also be 1.
All you can say is (p+n) = 5a + 1
Similarly, all you can say about statement 2 is (p-n) = 3b + 1

So using both also, you get (5a + 1)(3b + 1) = 15ab + 5a + 3b + 1
We know nothing about (5a + 3b + 1) - whether it is divisible by 15 or not.
_________________

Karishma
Veritas Prep GMAT Instructor

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Intern
Joined: 23 Oct 2012
Posts: 28

### Show Tags

14 Dec 2013, 00:18

I can see how based on your examples that my interpretation is not correct. However, i was going based off of a little tip in one of the other posts...

3) If a number leaves a remainder ‘r’ (the number is the divisor), all its factors will have the same remainder ‘r’ provided the value of ‘r’ is less than the value of the factor.
Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 7 (which is a factor of 21) will also be 5.

How is the situation in the problem different ...wherein the statement is not holding true!?
Intern
Joined: 23 Oct 2012
Posts: 28

### Show Tags

14 Dec 2013, 00:26
audiogal101 wrote:

I can see how based on your examples that my interpretation is not correct. However, i was going based off of a little tip in one of the other posts...

3) If a number leaves a remainder ‘r’ (the number is the divisor), all its factors will have the same remainder ‘r’ provided the value of ‘r’ is less than the value of the factor.
Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 7 (which is a factor of 21) will also be 5.

How is the situation in the problem different ...wherein the statement is not holding true!?

Ummm...i guess the statement is true...its just that 15 is not a factor of 5. Its the other way round.

So if the statement was (p+n) when divided by 15 gives a remainder of 1, then we can say (p+n) divided by 5 also gives a remainder of 1...
but if (p+n) leaves a remainder of 1 when divided by 5 , does not mean (p+n) divided by 15 will also leave remainder of 1...(as clearly evidenced by Karishma's example)
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8550
Location: Pune, India

### Show Tags

14 Dec 2013, 00:37
2
4
audiogal101 wrote:

I can see how based on your examples that my interpretation is not correct. However, i was going based off of a little tip in one of the other posts...

3) If a number leaves a remainder ‘r’ (the number is the divisor), all its factors will have the same remainder ‘r’ provided the value of ‘r’ is less than the value of the factor.
Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 7 (which is a factor of 21) will also be 5.

How is the situation in the problem different ...wherein the statement is not holding true!?

That is fine.

But this does not imply that if one factor leaves a remainder 'r', all multiples will leave remainder 'r' too. On the other hand, if all factors of that particular multiple leave the remainder 'r', then the multiple will leave remainder 'r' too.

Take an example:
If n is divided by 21 and it leaves remainder 1, when n is divided by 7, remainder will still be 1. When n is divided by 3, remainder will still be 1. - Correct

If n is divided by 7 and it leaves remainder 1, it doesn't mean that when n is divided by 14/21/28/35 ... it will leave remainder 1 in all cases.

But if n divided by 7 leaves remainder 1 and when divided by 3 leaves a remainder 1 too, it will leave remainder 1 when divided by 21 too.

For an explanation of these, check:
http://www.veritasprep.com/blog/2011/04 ... unraveled/
http://www.veritasprep.com/blog/2011/04 ... y-applied/
http://www.veritasprep.com/blog/2011/05 ... emainders/
_________________

Karishma
Veritas Prep GMAT Instructor

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Intern
Joined: 23 Oct 2012
Posts: 28

### Show Tags

14 Dec 2013, 01:01
VeritasPrepKarishma wrote:
audiogal101 wrote:

I can see how based on your examples that my interpretation is not correct. However, i was going based off of a little tip in one of the other posts...

3) If a number leaves a remainder ‘r’ (the number is the divisor), all its factors will have the same remainder ‘r’ provided the value of ‘r’ is less than the value of the factor.
Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 7 (which is a factor of 21) will also be 5.

How is the situation in the problem different ...wherein the statement is not holding true!?

That is fine.

But this does not imply that if one factor leaves a remainder 'r', all multiples will leave remainder 'r' too. On the other hand, if all factors of that particular multiple leave the remainder 'r', then the multiple will leave remainder 'r' too.

Take an example:
If n is divided by 21 and it leaves remainder 1, when n is divided by 7, remainder will still be 1. When n is divided by 3, remainder will still be 1. - Correct

If n is divided by 7 and it leaves remainder 1, it doesn't mean that when n is divided by 14/21/28/35 ... it will leave remainder 1 in all cases.

But if n divided by 7 leaves remainder 1 and when divided by 3 leaves a remainder 1 too, it will leave remainder 1 when divided by 21 too.

For an explanation of these, check:
http://www.veritasprep.com/blog/2011/04 ... unraveled/
http://www.veritasprep.com/blog/2011/04 ... y-applied/
http://www.veritasprep.com/blog/2011/05 ... emainders/

Thanks a ton!! You do an excellent job of deconstructing a complex issue and presenting it in simple and interesting ways...
SVP
Joined: 06 Sep 2013
Posts: 1745
Concentration: Finance

### Show Tags

27 Jan 2014, 17:18
1
VeritasPrepKarishma wrote:
audiogal101 wrote:

I can see how based on your examples that my interpretation is not correct. However, i was going based off of a little tip in one of the other posts...

3) If a number leaves a remainder ‘r’ (the number is the divisor), all its factors will have the same remainder ‘r’ provided the value of ‘r’ is less than the value of the factor.
Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 7 (which is a factor of 21) will also be 5.

How is the situation in the problem different ...wherein the statement is not holding true!?

That is fine.

But this does not imply that if one factor leaves a remainder 'r', all multiples will leave remainder 'r' too. On the other hand, if all factors of that particular multiple leave the remainder 'r', then the multiple will leave remainder 'r' too.

Take an example:
If n is divided by 21 and it leaves remainder 1, when n is divided by 7, remainder will still be 1. When n is divided by 3, remainder will still be 1. - Correct

If n is divided by 7 and it leaves remainder 1, it doesn't mean that when n is divided by 14/21/28/35 ... it will leave remainder 1 in all cases.

But if n divided by 7 leaves remainder 1 and when divided by 3 leaves a remainder 1 too, it will leave remainder 1 when divided by 21 too.

For an explanation of these, check:
http://www.veritasprep.com/blog/2011/04 ... unraveled/
http://www.veritasprep.com/blog/2011/04 ... y-applied/
http://www.veritasprep.com/blog/2011/05 ... emainders/

I personally found this a bit tricky

For instance:

RoF x*y / n = Rof (x/n) * Rof (y/n) / n

Now when we have 15 and we can divide it into 5*3. I'm trying to figure out what's wrong with this approach cause I did the following

Remainder of p+n/5 as per the first statement is 1

Remainder of p-n/3 as per the second statement is 1

Therefore, remainder of product of 1*1 / 15 is 1

I thought this was C

Any clues?
Cheers!
J
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8550
Location: Pune, India

### Show Tags

27 Jan 2014, 20:33
1
2
jlgdr wrote:
I personally found this a bit tricky

For instance:

RoF x*y / n = Rof (x/n) * Rof (y/n) / n

Now when we have 15 and we can divide it into 5*3. I'm trying to figure out what's wrong with this approach cause I did the following

Remainder of p+n/5 as per the first statement is 1

Remainder of p-n/3 as per the second statement is 1

Therefore, remainder of product of 1*1 / 15 is 1

I thought this was C

Any clues?
Cheers!
J

This would have been correct had the situation been:

"Remainder of(p+n)/15 as per the first statement is 1
Remainder of (p-n)/15 as per the second statement is 1
Therefore, remainder of product of 1*1 / 15 is 1"

But the statements tell us what happens when (p+n) or (p-n) is divided by 5 or 3 not 15.

Statement 1: p+n = 5a + 1
Statement 2: p-n = 3b + 1

(p+n)(p-n) = 15ab + 5a + 3b + 1
Only the first term is divisible by 15. We don't know the remainder when the rest of the 3 terms are divided by 15.

On the other hand, had the statements been a little different, your method would have been correct.
Statement 1: p+n = 15a + 1
Statement 2: p-n = 15b + 1
(p+n)(p-n) = 15*15ab + 15a + 15b + 1
Now the remainder would be 1.
_________________

Karishma
Veritas Prep GMAT Instructor

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Manager
Joined: 09 Nov 2013
Posts: 80
Re: If p and n are positive integers and p > n, what is the  [#permalink]

### Show Tags

02 Sep 2014, 05:57
VeritasPrepKarishma wrote:
jlgdr wrote:
I personally found this a bit tricky

For instance:

RoF x*y / n = Rof (x/n) * Rof (y/n) / n

Now when we have 15 and we can divide it into 5*3. I'm trying to figure out what's wrong with this approach cause I did the following

Remainder of p+n/5 as per the first statement is 1

Remainder of p-n/3 as per the second statement is 1

Therefore, remainder of product of 1*1 / 15 is 1

I thought this was C

Any clues?
Cheers!
J

This would have been correct had the situation been:

"Remainder of(p+n)/15 as per the first statement is 1
Remainder of (p-n)/15 as per the second statement is 1
Therefore, remainder of product of 1*1 / 15 is 1"

But the statements tell us what happens when (p+n) or (p-n) is divided by 5 or 3 not 15.

Statement 1: p+n = 5a + 1
Statement 2: p-n = 3b + 1

(p+n)(p-n) = 15ab + 5a + 3b + 1
Only the first term is divisible by 15. We don't know the remainder when the rest of the 3 terms are divided by 15.

On the other hand, had the statements been a little different, your method would have been correct.
Statement 1: p+n = 15a + 1
Statement 2: p-n = 15b + 1
(p+n)(p-n) = 15*15ab + 15a + 15b + 1
Now the remainder would be 1.

Intern
Joined: 02 Sep 2014
Posts: 15
Re: If p and n are positive integers and p > n, what is the  [#permalink]

### Show Tags

02 Sep 2014, 21:03
ok, i'll explain. by the way, the problem statement contains 'q' but the rest of everything contains 'n', so i'm going to go with 'n' just for the sake of consistency and because it's easier to type.the essence of this solution is that lionking is just finding different examples of numbers that satisfy the conditions, and just trying those numbers out.
Intern
Joined: 09 Oct 2014
Posts: 5
Re: If p and n are positive integers and p > n, what is the  [#permalink]

### Show Tags

17 Oct 2014, 12:15
BANON wrote:
If p and n are positive integers and p > n, what is the remainder when p^2 - n^2 is divided by 15 ?

(1) The remainder when p + n is divided by 5 is 1.
(2) The remainder when p - n is divided by 3 is 1.

i plugged in a 9 for p and a 2 for n.

i don't get the theory,,, so when i get confused i just try to solve it.
11/5 has a remainder of 1, i get other cans, so ns.

9-2 = 7= divided by 3 is a 1. granted i know others may have this.

but to have a remainder of one, and to be be dived by both a factor of 1 and 5, this fits. thus i plugged in the numbers for 7^2 = 49 - 2^2= 4 thus equals 45/15 = 0

i assumed that the remainder 15 was a factor of the 5 and 3. thus i picked c.

how in the world would i know not to do this?

thanks,.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8550
Location: Pune, India
Re: If p and n are positive integers and p > n, what is the  [#permalink]

### Show Tags

18 Oct 2014, 07:19
BANON wrote:
If p and n are positive integers and p > n, what is the remainder when p^2 - n^2 is divided by 15 ?

(1) The remainder when p + n is divided by 5 is 1.
(2) The remainder when p - n is divided by 3 is 1.

i plugged in a 9 for p and a 2 for n.

i don't get the theory,,, so when i get confused i just try to solve it.
11/5 has a remainder of 1, i get other cans, so ns.

9-2 = 7= divided by 3 is a 1. granted i know others may have this.

but to have a remainder of one, and to be be dived by both a factor of 1 and 5, this fits. thus i plugged in the numbers for 7^2 = 49 - 2^2= 4 thus equals 45/15 = 0

i assumed that the remainder 15 was a factor of the 5 and 3. thus i picked c.

how in the world would i know not to do this?

thanks,.

How do you know that for every suitable value of p and n, the remainder will always be the same?

Say p = 5 and n = 1.
p+n = 6 when divided by 5 gives remainder 1.
p-n = 4 when divided by 3 gives remainder 1.

p^2 - n^2 = 25 - 1 = 24
When this is divided by 15, remainder is 9. But you got remainder 0 with your values. Hence the remainder can take different values.

You cannot be sure that every time the remainder obtained will be 0. You cannot use number plugging to prove something. You cannot be sure that you have tried every possible value which could give a different result. You will need to rely on theory to prove that the remainder obtained will be the same in each case (or may not be the same). Using number plugging to disprove something is possible since all you have to do is find that one value for which the result does not hold.
_________________

Karishma
Veritas Prep GMAT Instructor

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8550
Location: Pune, India
Re: If p and n are positive integers and p > n, what is the  [#permalink]

### Show Tags

05 Jan 2015, 20:28
Quote:
If p and n are positive integers and p > n, what is the remainder when p^2 - n^2 is divided by 15 ?

(1) The remainder when p + n is divided by 5 is 1.
(2) The remainder when p - n is divided by 3 is 1.

You have explained the above question in one of the way. But if we think in the below way can u point out what is the mistake I am making?

p+n = 5t +1 : 1,6, 11, 16,21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76..........
p-n = 3m + 1 : 1, 4,7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79.......

Hence the general formula tht can be written is = 15 k + 1 & hence the remainder shall be 1 ??

Please let me where is the gap in the understanding??

Think on a few points:

The general formula 15k+1 is the formula for what? What number will be in this format?
Are you saying that the format of p^2 - n^2 will be this? If yes, then think - why?

p+n = 5t +1
p-n = 3m + 1

$$p^2 - n^2 = (p+n)(p-n) = (5t+1)(3m+1) = 15tm + 5t + 3m + 1$$

How did you get (15k+1)? How can we be sure that 5t+3m is divisible by 15?

The same thing was done by another reader above and I explained why this is wrong here: if-p-and-n-are-positive-integers-and-p-n-what-is-the-128002.html#p1304072
_________________

Karishma
Veritas Prep GMAT Instructor

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Director
Joined: 07 Aug 2011
Posts: 539
GMAT 1: 630 Q49 V27
Re: If p and n are positive integers and p > n, what is the  [#permalink]

### Show Tags

17 Mar 2015, 10:32
BANON wrote:
If p and n are positive integers and p > n, what is the remainder when p^2 - n^2 is divided by 15 ?

(1) The remainder when p + n is divided by 5 is 1.
(2) The remainder when p - n is divided by 3 is 1.

Option A:
At P=5 N=1 P+N when divided by 5 will have remainder 1 AND $$P^2-N^2 / 15$$ will have remainder 9.
At P=15 N=1 P+N when divided by 5 will have remainder 1 AND $$P^2-N^2 / 15$$ will have remainder 14.

Option B:
At P=5 N=1 P-N when divided by 3 will have remainder 1 AND $$P^2-N^2 / 15$$ will have remainder 9.
At P=6 N=2 P-N when divided by 3 will have remainder 1 AND $$P^2-N^2 / 15$$ will have remainder 2.

Both A and B are not sufficient.

Combining
P+N = 5K + 1 i.e. and ODD Integer
P-N = 3X+1 i.e. an EVEN Integer and so it has no say on the divisibility by 15 it acts as a constant in P^2-N^2.
(P+N) will decided whether $$P^2 - N^2$$ is divisible by 15 and also the remainder , but we have already proved in option A that P+N can have two different remainders (9,14 , we can test more numbers) and also (P-N) when multiplied to these two remainders will definitely give different remainders when divided by 15.

_________________

Thanks,
Lucky

_______________________________________________________
Kindly press the to appreciate my post !!

Senior Manager
Joined: 08 Dec 2015
Posts: 294
GMAT 1: 600 Q44 V27
Re: If p and n are positive integers and p > n, what is the  [#permalink]

### Show Tags

07 Jul 2016, 02:51
Am I underestimating this question?

For me when taking both 1 and 2:

We know that p+n=5q+1 and p-n=3q+1

well just plug the same value for each q and get: for q=1 (6*4) and just divide it by 15 to get R9. then plug q=2 and get that 77/15 has R2 so that's it, end of story, insufficient.

Am I missing something?
Re: If p and n are positive integers and p > n, what is the &nbs [#permalink] 07 Jul 2016, 02:51

Go to page    1   2    Next  [ 27 posts ]

Display posts from previous: Sort by