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

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.
_________________

I'm happy to help with it here.

The first skill we need is the "rebuilding the dividend" equation. See:
https://magoosh.com/gmat/2012/gmat-quant ... emainders/
If we divide N by D, and get a quotient of Q with remainder of R, then
N/D = Q + (R/D)
or
N = Q*D + R
That is one of the most powerful and unappreciated formulas on the GMAT, the "rebuilding the dividend" equation. As for the name, the terminology here is
N = the dividend, the thing divided
D = the divisor, the number by which we are dividing
Q = the quotient, the result of division
R = the remainder

Here, we can change the statements into equations:
Statement #1: (p + n) = 5*S + 1, for some quotient integer S
Statement #2: (p - n) = 3*T + 1, for some quotient integer T

The second thing that you really need to recognize quickly on the GMAT is the Difference of Two Squares patterns, the single most important algebraic pattern on the entire GMAT. See:
https://magoosh.com/gmat/2012/gmat-quant ... o-squares/
https://magoosh.com/gmat/2013/three-alge ... -the-gmat/
Here, this implies that:
p^2 - n^2 = (p + n)*(p - n)

One statement talks only about 5, and the other, only about 3, so neither is sufficient individually. If both are considered together, we can multiply the two formulas together:
p^2 - n^2 = (p + n)*(p - n) = (5*S + 1)*(3*T + 1) = 15*ST + 5*S + 3*T + 1
Well, clearly the first term, (15*ST) would be divisible by 15. The problem is: we don't know the values of S or T. We know they are integers, but we don't know their values. If think about possible values of 5*S, multiples of 5, then when we divide multiples of 5 by 15, we could have remainders of 0, 5, or 10. Similarly, when we divided 3*T, multiples of 3, by 15, we could have remainders of 0, 3, 6, 9, or 12. Thus, 15 will go evenly into the first term, but may have a remainder on the second term and on the third time, and of course, will have a remainder of 1 on the last term. Because there's still choice about the possible remainders, even with both statements combined, we cannot answer the prompt question.

Even with both statements combined, everything is still insufficient. Answer = (E)

Does all this make sense?
Mike
_________________

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.
_________________

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.
_________________

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://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html/2011/04 ... unraveled/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html/2011/04 ... y-applied/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html/2011/05 ... emainders/
_________________

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.
_________________

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
_________________

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.
_________________

@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.
_________________

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.
_________________

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.
_________________

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.
_________________