joannabanana 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.

p^2 - n^2 = (p+n)(p-n). So the question can be rewritten as:

What is the remainder when (p+n)(p-n) is divided by 15?

Statement 1:
Tells us nothing about p-n. Insufficient.

Statement 2:
Tells us nothing about p+n. Insufficient.

Statements 1 and 2 together:
The remainder when p+n is divided by 5 is 1. This means that p+n is a (multiple of 5) + 1. Multiples of 5 are 0,5,10,15,20, etc.
So p+n = 1,6,11,16,21,etc.

The remainder when p-n is divided by 3 is 1. This means that p-n is a (multiple of 3) + 1. Multiples of 3 are 0,3,6,9,12, etc.
So p-n = 1,4,7,10,13,etc.

If p+n=11 and p-n=1 so that p=6 and n=5, then (p+n)(p-n) = 11*1 = 11.
11/15 = 0 R11.
If p+n=21 and p-n=1 so that p=11 and n=10, then (p+n)(p-n) = 21*1 = 21.
21/15 = 1 R6.

Since the remainder can take on different values, insufficient.

The correct answer is E.

Factor p^2-n^2 into (p+n)(p-n), and see if you can come up with more than one possible remainder when dividing this product by 15. The key to ease your plugging in is to remember that when dividing a smaller number by a larger number, the remainder is always equal to the smaller number itself. e.g. When dividing 6 by 15, the quotient is zero, and the remainder is 6. Thus, no need to stick to numbers greater than 15 for the entire problem, a process which will make you sorry you were born.

Stat. (1): p+n can be such values as 6, 11, 16 - all of which give a remainder of 1 when divided by 5. Even if we discount the additional factor of p-n (we can always choose such values of p and n so that p-n=1), these values alone give different remainders when divided by 15: 6 /15 gives a remainder of 6, 11/15 gives a remainder of 11, 16/15 gives a remainder of 1. Insuff.

Stat. (2): p-n can be 1, 4, 7. My intuition is again to discount p+n here, but there's an added complication that p and n must be positive, meaning they cannot equal 1. Still: take p=3 and n=2 (so that p-n=1 and p+n=5) to get 1*5=24 with a remainder of 5 when div. by 15, and p=6 and n=2 to get (so that p-n=4 and p+n=8) to get 4*8=32 with a remainder of 2 when div. by 15. Insuff.

Combined: look for examples that satisfy both: p-n=1 and p+n=11 (6 +/- 5), or p-n=7, p+n=11 (9 +/- 2). In the first case, the expression in the question stem is 11 with a remainder of 11; in the second case, the expression is 7*11=77, with a remainder of 2. Insuff.

Thanks . Right strategy .

Could someone also explain algebric way to solve this, not number picking ?