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

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Joined: 02 Sep 2009
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Re: If p and n are positive integers and p > n, what is the  [#permalink]

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07 Jul 2016, 04:31
iliavko wrote:
Am I underestimating this question?

For me when taking both 1 and 2:

We know that p+n=5q+1 and p-n=3q+1

well just plug the same value for each q and get: for q=1 (6*4) and just divide it by 15 to get R9. then plug q=2 and get that 77/15 has R2 so that's it, end of story, insufficient.

Am I missing something?

No, that's not correct. You cannot put the same quotient in both cases. Each solution on previous page makes this clear.
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Re: If p and n are positive integers and p > n, what is the  [#permalink]

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13 Feb 2018, 20:44
Bunuel wrote:
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.

Hi Bunuel

I plugged in numbers

So i had chosen 31 which is divisble by 3 and 5. Hence, i chose C

Could u pls explain what im missing out on?
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Re: If p and n are positive integers and p > n, what is the  [#permalink]

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13 Feb 2018, 20:53
zanaik89 wrote:
Bunuel wrote:
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.

Hi Bunuel

I plugged in numbers

So i had chosen 31 which is divisble by 3 and 5. Hence, i chose C

Could u pls explain what im missing out on?

One example is not enough to get sufficiency. Also, when asking a question, please be more specific and show you work.
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Re: If p and n are positive integers and p > n, what is the  [#permalink]

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17 Feb 2018, 23:09
BANON 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.

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.

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.
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