p and n are positive integers

[suchi]

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

The remainder when p+q is divided by 5 is 1
The remainder when p-q is divided by 3 is 1

[lionkiNg]

Use listing method..

p+n=11
p-n=1

R=11

p+n=6
p-n=1

R=6

So answer is E.

[RonPurewal]

Use listing method..

p+n=11
p-n=1

R=11

p+n=6
p-n=1

R=6

So answer is E.

good enough.

remember, if you aren't getting anywhere by using theory, you should IMMEDIATELY start using secondary methods (such as listing possibilities). it can sometimes take a relatively long time to find the possibilities you need, so you shouldn't delay. do not deliberate.

[so19]

However there are no positive integer p and n that satisfy the below example.

p+n=6
p-n=1

p = 7/2, n = 5/2
These are not positive integers!

Therefore
p+n=6
p-n=1
is a bad example.

p+n=16
p-n=4
works better since r = 4 and (p,n) = (10,6).

[amit]

Use listing method..

p+n=11
p-n=1

R=11

p+n=6
p-n=1

R=6

So answer is E.

I don't understand a thing :mrgreen: ! Please explain

[RonPurewal]

Use listing method..

p+n=11
p-n=1

R=11

p+n=6
p-n=1

R=6

So answer is E.

I don't understand a thing :mrgreen: ! Please explain

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.

essential fact: this problem features the DIFFERENCE OF SQUARES, p^2 - n^2 = (p - n)(p + n). if you don't make this realization IMMEDIATELY, then you're going to be mired forever in trying to solve for p and n individually. (remember that, if a problem is stated entirely in terms of combinations, then the problem can almost certainly be solved entirely in terms of combinations.)

lionking's first example chooses p + n = 11 (which gives a remainder of 1 upon division by 5, as required), and p - n = 1 (which gives a remainder of 1 upon division by 3, as required).
when 11 x 1 = 11 is divided by 15, the remainder is 11.

lionking's second example chooses p + n = 6 (which gives a remainder of 1 upon division by 5, as required), and p - n = 1 (which gives a remainder of 1 upon division by 3, as required).
when 6 x 1 = 11 is divided by 15, the remainder is 6.

these are two different remainders, so, insufficient.

[chandrahasreddy]

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.

essential fact: this problem features the DIFFERENCE OF SQUARES, p^2 - n^2 = (p - n)(p + n). if you don't make this realization IMMEDIATELY, then you're going to be mired forever in trying to solve for p and n individually. (remember that, if a problem is stated entirely in terms of combinations, then the problem can almost certainly be solved entirely in terms of combinations.)

lionking's first example chooses p + n = 11 (which gives a remainder of 1 upon division by 5, as required), and p - n = 1 (which gives a remainder of 1 upon division by 3, as required).
when 11 x 1 = 11 is divided by 15, the remainder is 11.

lionking's second example chooses p + n = 6 (which gives a remainder of 1 upon division by 5, as required), and p - n = 1 (which gives a remainder of 1 upon division by 3, as required).
when 6 x 1 = 11 is divided by 15, the remainder is 6.

these are two different remainders, so, insufficient.

I don't mean to rain on the parade...but
p + n = 6
p - n = 1 do not yield integers for p and n.(that is required as per problem)

I guess we can choose ..for the second case
p + n = 6
p - n = 4
and we get- insufficient.

[RonPurewal]

chandra, good catch.