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If p and n are positive integers and p > n, what is the remainder when [#permalink]
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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:
https://youtu.be/A5abKfUBFSc
https://anaprep.com/number-properties-b ... isibility/
https://anaprep.com/number-properties-b ... isibility/
https://anaprep.com/number-properties-m ... emainders/
https://anaprep.com/number-properties-a ... -shortcut/
https://anaprep.com/number-properties-t ... emainders/

Originally posted by KarishmaB on 14 Dec 2013, 01:37.
Last edited by KarishmaB on 21 Dec 2023, 07:31, edited 1 time in total.
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Re: If p and n are positive integers and p > n, what is the remainder when [#permalink]
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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.

Answer: E.


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.
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Re: If p and n are positive integers and p > n, what is the remainder when [#permalink]
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Giorgosaek wrote:
Hi experts Bunuel, VeritasKarishma, @chetan2u

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 thoughts?


Giorgosaek

What you said works when we work with the same dividend.

A/5 Rem 1
A/3 Rem 1
Then A/15 gives Rem 1
It works because of the logic discussed here:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... emainders/

Here the two dividends are not same. Hence, this will not work.
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Re: If p and n are positive integers and p > n, what is the remainder when [#permalink]
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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 ?

p^2 - n^2 = (p+n) (p-n)
Here we need to find the remainder when (p+n)(p-n) is divided by 15

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

The only thing we can conclude here is p+n = 5 m + 1 , where m is a constant. Also we don't have any info about p-n.
Hence Statement 1 is insufficient.

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

p-n = 3n + 1
By using statement 2 alone , we don't have any idea about p+n.
Hence its insufficient.

Lets try to combine both statements and see whether we can find the answer or not .

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

(p+n)(p-n) = (5 m + 1)(3n + 1)
= 15 mn + 5m + 3n + 1

When you divide (p+n)(p-n) by 15 , the remainder will depend on value of m and n
Lets try with an example:
if m= 1, n =1
p+ n = 6
p-n = 4

p=5, n= 1 so the condition p> n is satisfied.

(p+n)(p-n) = 6*4 = 24

Remainder when 24 is divided by 15 is 9

Lets try another example:

if m= 2, n =2
p+ n = 11
p-n = 7

p=9, n= 2 so the condition p> n is satisfied.

(p+n)(p-n) = 11*7 = 77

Remainder when 77 is divided by 15 is 2

Since you are not able to get a definite remainder, Option E would be the correct answer.

Thanks,
Clifin J Francis
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Re: If p and n are positive integers and p > n, what is the remainder when [#permalink]
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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.

Answer: E.


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
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Re: If p and n are positive integers and p > n, what is the remainder when [#permalink]
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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?

From your work,
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
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Re: If p and n are positive integers and p > n, what is the remainder when [#permalink]
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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.


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (p and n) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first.

Conditions 1) & 2):
\(p + n = 6, p - n = 4 : p = 5, n = 1 : p^2 - n^2 = 25 - 1 = 24\), its remainder is \(4\).
\(p + n = 11, p - n = 7 : p = 9, n = 2 : p^2 - n^2 = 81 - 4 = 77\), its remainder is \(2\).

Thus, both conditions together are not sufficient, since the remainder is not unique.

Therefore, E is the answer.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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Re: If p and n are positive integers and p > n, what is the remainder when [#permalink]
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aadilismail wrote:
@VertiasKarishma
once we've narrowed down to C/E
can we test the relationship in a simultaneous equation setup by adding p+n = 5Q+1 and p-n = 3K+1 which
works out to 2p = 5Q + 3K + 2 i.e. three variables requiring three assumptions hence unsolvable ?
Thus E is the best answer.


No, this does not work out. There is no point looking for p. We need p^2 - n^2. Sometimes, you need less information to get the value of what you need even though you may not be able to get the value of each variable independently.

p + n = 5Q + 1

p - n = 3K + 1

Now, what you need is (p + n)*(p - n)

(p + n)*(p - n) = (5Q + 1)*(3K + 1) = 15QK + 3K + 5Q + 1

Is this divisible by 15? We know that the first term (15QK) is but what about the sum of the other 3 terms? We don't know since that depends on the values of Q and K.
If K = 5 and Q = 3, this is not divisible by 15.
If K = 3 and Q = 4, this is divisible by 15.
Hence, both statements are insufficient.
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Re: If p and n are positive integers and p > n, what is the remainder when [#permalink]
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singh8891 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.

Answer: E.


Hello
When the question stem says that The remainder when p + n is divided by 5 is 1, doesn't it implicitly place a constraint on p-n? So dont we need to check p-n too?
While solving the question, I calculated p+n based on the criteria, p-n for the chosen numbers and then p^2-n^2 and finally checked divisibility. I started running out of time and had to guess and move.


Note the information given to you about (p + n). The remainder when you divide it by 5 is 1.

So (p+n) can be 6/11/16/21/26/31/36... etc. This gives us limitless values for (p-n) with all possible different remainders when divided by 5.

Yes, a particular small value for (p+n) could have put a constraint on (p - n). Such as (p+n) = 6. Now (p - n) could take a few values only.
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If p and n are positive integers and p > n, what is the remainder when [#permalink]
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kagrawal16 wrote:
Hello VeritasKarishma,

I did this question by reasoning of a number properties concept. Can you please check if the method is correct as I am not sure. I did get the approach in the solutions though.

1) p+n is not divisible by 5.
implies p or n not divisible by 5 OR both not divisible by 5 (I mean for eg. p=7 & n=2 then p+n = 9 but p-n can be 5)
Therefore p-n cannot be divisible by 5 OR p-n can be divisible by 5.
In such a case p-n can be divisible by 3 too. So out.

2) Same logic as 1.

Combining
p-n can be divisible by 5 OR p+n can be divisible by 3. if Both are true then yes, else no.

Also, I suppose the approach is a bit narrow as the question is "what is the remainder", whereas i checked whether was there a remainder=0 or not.


I found it hard to follow your train of thought. It will be better if you take a more structured approach as shown by Bunuel in this post: https://gmatclub.com/forum/if-p-and-n-a ... l#p1048742

When you get 15kt + 5t + 3k + 1, you know that t = 1, k = 1 will give you 9 remainder. But t = 2, k = 1 will give you 14 remainder. So you cannot find a unique value for remainder.
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Re: If p and n are positive integers and p > n, what is the remainder when [#permalink]
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rsrighosh wrote:
if given \(\frac{p+n }{ 5}\) gives remainder as \(1\), doesnt it also mean \(\frac{p-n}{5}\) also should give remainder as \(1\) ?

VeritasKarishma chetan2u Brian123 Bunuel


No, it will not be the case.

Say p=n
p-n will be 0 and so divisible by all numbers including 5.
But p+n =2p and divisibility will depend on value of p
p=3, 3+3 will leave remainder 1.
p=4, 4+4 will leave 3
p=5, 5+5 will be divisible by 5.
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Re: If p and n are positive integers and p > n, what is the remainder when [#permalink]
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Hi

Always remember that the remainders also get added and subtracted as you add different numbers

So let p and n when divided by 5 give remainder 4 and 2, then p+n will give remainder 4+2 or 6 or 1.
But p-n will give remainder 4-2 or 2.

Choose numbers accordingly 14 and 12...14+12=26, so remainder is 1, but 14-12=2.
There may be cases when it is same.
For example 5 and 6....6+5 and 6-5 both will give 1.

So we cannot be sure what remainder is.
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Re: If p and n are positive integers and p > n, what is the remainder when [#permalink]
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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.

Hello, everyone. I came across this question today in a session with a student and used a number-sense approach to disqualify any hopes of pinning down an answer. I can see that others have started along the same lines—spotting the difference of squares, brainstorming numbers—but not quite pursued the same line of reasoning thereafter, so I thought I would share my approach, just in case it may help someone else. I am going to pick up my analysis from the (C) versus (E) perspective:

Statement (1):

First, (p + n) CANNOT be 1 because p and n must EACH be positive integers (0 does not qualify). So, we are looking at 6, 11, 16, 21, 26, 31, 36, 41, 46, 51... (just as GMATGuruNY has outlined in an earlier post).

Statement (2):

(p - n) CAN be 1 because two positive integers could conceivably produce such a difference. So, we are looking at 1, 4, 7, 10, 13, 16, 19, 22, 25, 28...

All we have to do is figure out valid combinations of positive integers that could both satisfy the statements and conform to the information given in the problem. Start small:

(p + n) = 6
LET p = 5 and n = 1.
p > n

(p - n) = (5 - 1) = 4
This value appears in the second list of numbers from above, so p = 5, n = 1 is a VALID combination to test within the question.

\(\frac{(p^2 - n^2)}{15}\)

\(\frac{(5^2 - 1^2)}{15}\)

\(\frac{(25 - 1)}{15}\)

\(\frac{24}{15}\)

The remainder is 9.

Now, our task is to check whether any other valid combinations of p and n values will yield the same remainder. We can even go back to (p + n) = 6 and check other values.

LET p = 4 and n = 2.
p > n

(p - n) = (4 - 2) = 2 X

Since 2 is NOT in the second list of numbers from earlier, we can disqualify this combination of values. Furthermore, we CANNOT use 3 and 3 since they would not satisfy the given inequality p > n. We need to test different numbers. How about 11, the next number up, in our sum? We have 10 + 1, 9 + 2, 8 + 3, 7 + 4, and 6 + 5 to run through, potentially.

LET p = 10 and n = 1.
p > n

(p - n) = (10 - 1) = 9 X

LET p = 9 and n = 2.
p > n

(p - n) = (9 - 2) = 7

We have another VALID combination of values to test. Go back to the question and substitute:

\(\frac{(p^2 - n^2)}{15}\)

\(\frac{(9^2 - 2^2)}{15}\)

\(\frac{(81 - 4)}{15}\)

\(\frac{77}{15}\)

The remainder is 2.

Now, you can see that we CANNOT tell what the remainder will be, since we have created two perfectly valid cases that produced different remainders. Thus, the answer must be (E).

You might see the above approach as sloppy or less refined than the methods outlined above, but it proved pretty efficient in my case, and I was certain I had the answer. (And trust me, GMAC™ does not reward bonus points for elegant work.)

Good luck with your studies, however you tackle these tough Quant questions.

- Andrew
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Re: If p and n are positive integers and p > n, what is the remainder when [#permalink]
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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.
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Re: If p and n are positive integers and p > n, what is the remainder when [#permalink]
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: If p and n are positive integers and p > n, what is the remainder when [#permalink]
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,.
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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.

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.
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