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remainders

by aleph777 » Tue Feb 01, 2011 11:48 am
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.

Answer: E

I answered C, thinking you could multiply the two statements together to solve for the initial question (after breaking down p^2 - n^2 into (p + n)(p - n). Clearly, I was wrong, but I want to understand the rationale.

Can you only combine remainder equations when they have the same denominator?

Thanks!
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by GMATGuruNY » Tue Feb 01, 2011 12:33 pm
aleph777 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.

Answer: E

I answered C, thinking you could multiply the two statements together to solve for the initial question (after breaking down p^2 - n^2 into (p + n)(p - n). Clearly, I was wrong, but I want to understand the rationale.

Can you only combine remainder equations when they have the same denominator?

Thanks!
When the statements are combined, p+n = (multiple of 5) + 1 and p-n = (multiple of 3) + 1.
We might be tempted to conclude that (p+n)(p-n) = (multiple of 15) + 1. Not so.

Let's look at the potential values of p+n and p-n:

p+n = 6, 11, 16, 21, 26, 31, 36, 41...
p-n = 1, 4, 7, 10...

The values of p+n are not all multiples of any common factor (other than 1).
The values of p-n are not all multiples of any common factor (other than 1).

So (p+n)(p-n) need not be a multiple of any particular factor. Thus, when (p+n)(p-n) is divided by 15, the remainder will take on different values.

The safest and quickest approach would be to write out lists for p+n and p-n (as shown above) and try out some cases:
Let p=6, n=5, so that p+n=11 and p-n=1.
Then (p+n)(p-n) = 11*1 = 11
11/15 = 0 R11.

Let p=11, n=10, so that p+n=21 and p-n=1.
Then (p+n)(p-n) = 21*1 = 21
21/15 = 1 R6.

Since R=11 and R=6, insufficient.

The correct answer is E.
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by Night reader » Tue Feb 01, 2011 1:18 pm
aleph777 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.

Answer: E

I answered C, thinking you could multiply the two statements together to solve for the initial question (after breaking down p^2 - n^2 into (p + n)(p - n). Clearly, I was wrong, but I want to understand the rationale.

Can you only combine remainder equations when they have the same denominator?

Thanks!
p>n {p,n - integers}
p^2-n^2=15i+r {i - integer, r - remainder}, find r-?

(1) p+n=5i+r Not Sufficient, as p-n=(15i+r)/(5i+1) three variables one equation;
(2) p-n=3i+r Not Sufficient, as p+n=(15i+r)/(3i+1) three variables one equation;
Combining st(1&2) (5i+r)(3i+r)=15i+r, 15i^2+5ir+3ir+r^2-15i-r=0, 15i(i-1)=r-8ir-r^2 If we can solve it for r then Sufficient, otherwise Not Sufficient

E
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by aleph777 » Tue Feb 01, 2011 2:36 pm
Pretty simple when you just plug in that way. Thanks.

GMATGuruNY wrote:
aleph777 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.

Answer: E

I answered C, thinking you could multiply the two statements together to solve for the initial question (after breaking down p^2 - n^2 into (p + n)(p - n). Clearly, I was wrong, but I want to understand the rationale.

Can you only combine remainder equations when they have the same denominator?

Thanks!
When the statements are combined, p+n = (multiple of 5) + 1 and p-n = (multiple of 3) + 1.
We might be tempted to conclude that (p+n)(p-n) = (multiple of 15) + 1. Not so.

Let's look at the potential values of p+n and p-n:

p+n = 6, 11, 16, 21, 26, 31, 36, 41...
p-n = 1, 4, 7, 10...

The values of p+n are not all multiples of any common factor (other than 1).
The values of p-n are not all multiples of any common factor (other than 1).

So (p+n)(p-n) need not be a multiple of any particular factor. Thus, when (p+n)(p-n) is divided by 15, the remainder will take on different values.

The safest and quickest approach would be to write out lists for p+n and p-n (as shown above) and try out some cases:
Let p=6, n=5, so that p+n=11 and p-n=1.
Then (p+n)(p-n) = 11*1 = 11
11/15 = 0 R11.

Let p=11, n=10, so that p+n=21 and p-n=1.
Then (p+n)(p-n) = 21*1 = 21
21/15 = 1 R6.

Since R=11 and R=6, insufficient.

The correct answer is E.
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by prachich1987 » Thu Feb 03, 2011 2:33 am
What if we changed the second statement to below


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

I think in this case the reminder would be 1.Right?
Thanks!
Prachi
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