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

chetan2u But the question explictly says p>n. So p not equal to n. So in that case will if p+n/5 gives remainder 1, will p-n/5 give remainder 1?

My question is that, if the above is correct, wont the answer be C?

I solved in this way :-

if statement 1 :- \(\frac{p+n}{5}\) gives remainder 1, will \(\frac{p−n}{5}\) give remainder 1. But we do not know whether (p+n)(p-n) is divisible by 3
if statement 2 :- \(\frac{ p-n}{3}\) gives remainder 1, will \(\frac{p+n}{3}\) give remainder 1. But we do not know whether (p+n)(p-n) is divisible by 5

Together statement 1 and 2 ---> (p+n) (p-n) will give remainder 16.. Hence C



Or my understanding is wrong here. I am not able to understand why answer is E if the statement above is true. Can you please help?
I pasted the question below as reference.

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.

Video solution from Quant Reasoning starts at 11:00
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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

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
GMAT SME

I tried below way which yielded incorrect answer. Could someone please help me explain which logic in incorrect here?

(p+n)(p-n)/15 - Need to find Remainder for this.

(p+n)(p-n)/5*3 => (p+n)/5 * (p-n)/3 => Remainder for first term * Remainder of 2nd term => 1* 1 => 1 is the answer.

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