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If p and n are positive integers and p > n, what is the
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23 Feb 2012, 07:10
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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.
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Re: DATA5
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23 Feb 2012, 07:20
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 \(p^2  n^2=(p+n)(pn)\). (1) The remainder when p + n is divided by 5 is 1. No info about pn. 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 \(pn=3k+1\). Multiply these two > \((p+n)(pn)=(5t+1)(3k+1)=15kt+5t+3k+1\), 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 \(pn=1\) (1 divided by 3 yields remainder of 1) then \((p+n)(pn)=11\) and remainder upon division 11 by 15 is 11 BUT if \(p+n=21\) (21 divided by 5 yields remainder of 1) and \(pn=1\) (1 divided by 3 yields remainder of 1) then \((p+n)(pn)=21\) and remainder upon division 21 by 15 is 6. Not sufficient. Answer: E.
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Re: DATA5
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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 \(p^2  n^2=(p+n)(pn)\).
(1) The remainder when p + n is divided by 5 is 1. No info about pn. 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 \(pn=3k+1\).
Multiply these two > \((p+n)(pn)=(5t+1)(3k+1)=15kt+5t+3k+1\), 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 \(pn=1\) (1 divided by 3 yields remainder of 1) then \((p+n)(pn)=11\) and remainder upon division 11 by 15 is 11 BUT if \(p+n=21\) (21 divided by 5 yields remainder of 1) and \(pn=1\) (1 divided by 3 yields remainder of 1) then \((p+n)(pn)=21\) and remainder upon division 21 by 15 is 6. Not sufficient.
Answer: E. Hi Bunuel  1 doubt.. why can't the below process be followed? p+n = 5A+1 => 1,6,11, 16,21,26 pn = 3B+1 => 1,4,7,10,13, 16,19,21 p+n * pn => 15 K + 16. Hence the remainder on division by 15 gives 1. Cheers



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Re: DATA5
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05 Nov 2012, 03:33
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 \(p^2  n^2=(p+n)(pn)\).
(1) The remainder when p + n is divided by 5 is 1. No info about pn. 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 \(pn=3k+1\).
Multiply these two > \((p+n)(pn)=(5t+1)(3k+1)=15kt+5t+3k+1\), 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 \(pn=1\) (1 divided by 3 yields remainder of 1) then \((p+n)(pn)=11\) and remainder upon division 11 by 15 is 11 BUT if \(p+n=21\) (21 divided by 5 yields remainder of 1) and \(pn=1\) (1 divided by 3 yields remainder of 1) then \((p+n)(pn)=21\) and remainder upon division 21 by 15 is 6. Not sufficient.
Answer: E. Hi Bunuel  1 doubt.. why can't the below process be followed? p+n = 5A+1 => 1,6,11, 16,21,26 pn = 3B+1 => 1,4,7,10,13, 16,19,21 p+n * pn => 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 pn= 4 remainder after dividing by 3 is 1 p^2 N2 = 251 = 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 pn = 7 remainder after dividing by 3 is 1 p^2  N^2 = 814 = 77 remainder after dividing by 15 is 2. hence both statement combined are not suff
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Re: DATA5
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06 Nov 2012, 04:21
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 \(p^2  n^2=(p+n)(pn)\).
(1) The remainder when p + n is divided by 5 is 1. No info about pn. 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 \(pn=3k+1\).
Multiply these two > \((p+n)(pn)=(5t+1)(3k+1)=15kt+5t+3k+1\), 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 \(pn=1\) (1 divided by 3 yields remainder of 1) then \((p+n)(pn)=11\) and remainder upon division 11 by 15 is 11 BUT if \(p+n=21\) (21 divided by 5 yields remainder of 1) and \(pn=1\) (1 divided by 3 yields remainder of 1) then \((p+n)(pn)=21\) and remainder upon division 21 by 15 is 6. Not sufficient.
Answer: E. Hi Bunuel  1 doubt.. why can't the below process be followed? p+n = 5A+1 => 1,6,11, 16,21,26 pn = 3B+1 => 1,4,7,10,13, 16,19,21 p+n * pn => 15 K + 16. Hence the remainder on division by 15 gives 1. Cheers p+n = 5A+1 and pn = 3B+1 does not mean that (p+n)*(pn)=15K+16. When you expand (p+n)*(pn)=(5A+1)(3B+1) you won't get an expression of the form 15K+16.
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Re: DATA5
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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 \(p^2  n^2=(p+n)(pn)\).
(1) The remainder when p + n is divided by 5 is 1. No info about pn. 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 \(pn=3k+1\).
Multiply these two > \((p+n)(pn)=(5t+1)(3k+1)=15kt+5t+3k+1\), 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 \(pn=1\) (1 divided by 3 yields remainder of 1) then \((p+n)(pn)=11\) and remainder upon division 11 by 15 is 11 BUT if \(p+n=21\) (21 divided by 5 yields remainder of 1) and \(pn=1\) (1 divided by 3 yields remainder of 1) then \((p+n)(pn)=21\) and remainder upon division 21 by 15 is 6. Not sufficient.
Answer: E. Hi Bunuel, I reasoned this question as follows: Prompt is asking what is remainder when... (p+n) (pn) /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 pn is divided by 3 is 1...this implies..remainder when (pn) 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 (pn) /15 gives remainder 1.....so (p+n) (pn)/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?



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Re: DATA5
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14 Dec 2013, 00:03
audiogal101 wrote: Prompt is asking what is remainder when... (p+n) (pn) /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 pn is divided by 3 is 1...this implies..remainder when (pn) 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 (pn) /15 gives remainder 1.....so (p+n) (pn)/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 (pn) = 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.
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Re: DATA5
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14 Dec 2013, 00:18
Thanks for your response Karishma.
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!?



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Re: DATA5
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14 Dec 2013, 00:26
audiogal101 wrote: Thanks for your response Karishma.
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)



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Re: DATA5
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14 Dec 2013, 00:37
audiogal101 wrote: Thanks for your response Karishma.
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 ... yapplied/http://www.veritasprep.com/blog/2011/05 ... emainders/
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Re: DATA5
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14 Dec 2013, 01:01
VeritasPrepKarishma wrote: audiogal101 wrote: Thanks for your response Karishma.
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 ... yapplied/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...



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Re: DATA5
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27 Jan 2014, 17:18
VeritasPrepKarishma wrote: audiogal101 wrote: Thanks for your response Karishma.
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 ... yapplied/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 pn/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



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Re: DATA5
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27 Jan 2014, 20:33
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 pn/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 (pn)/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 (pn) is divided by 5 or 3 not 15. Statement 1: p+n = 5a + 1 Statement 2: pn = 3b + 1 (p+n)(pn) = 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: pn = 15b + 1 (p+n)(pn) = 15*15ab + 15a + 15b + 1 Now the remainder would be 1.
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Re: If p and n are positive integers and p > n, what is the
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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 pn/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 (pn)/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 (pn) is divided by 5 or 3 not 15. Statement 1: p+n = 5a + 1 Statement 2: pn = 3b + 1 (p+n)(pn) = 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: pn = 15b + 1 (p+n)(pn) = 15*15ab + 15a + 15b + 1 Now the remainder would be 1. hey karishma! after reading your all informative articles, I am curious to know your GMAT score.



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Re: If p and n are positive integers and p > n, what is the
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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.



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Re: If p and n are positive integers and p > n, what is the
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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. 92 = 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,.



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Re: If p and n are positive integers and p > n, what is the
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18 Oct 2014, 07:19
ctkadmin wrote: 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. 92 = 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. pn = 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.
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Re: If p and n are positive integers and p > n, what is the
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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.......... pn = 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? From your work, p+n = 5t +1 pn = 3m + 1 \(p^2  n^2 = (p+n)(pn) = (5t+1)(3m+1) = 15tm + 5t + 3m + 1\) 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: ifpandnarepositiveintegersandpnwhatisthe128002.html#p1304072
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Re: If p and n are positive integers and p > n, what is the
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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 \(P^2N^2 / 15\) will have remainder 9. At P=15 N=1 P+N when divided by 5 will have remainder 1 AND \(P^2N^2 / 15\) will have remainder 14. Option B: At P=5 N=1 PN when divided by 3 will have remainder 1 AND \(P^2N^2 / 15\) will have remainder 9. At P=6 N=2 PN when divided by 3 will have remainder 1 AND \(P^2N^2 / 15\) will have remainder 2. Both A and B are not sufficient. Combining P+N = 5K + 1 i.e. and ODD Integer PN = 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^2N^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 (PN) when multiplied to these two remainders will definitely give different remainders when divided by 15. Answer : E
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Re: If p and n are positive integers and p > n, what is the
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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 pn=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
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