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# If p is a positive integer and r is the remainder when p2 1 is divide

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If p is a positive integer and r is the remainder when p2 1 is divide  [#permalink]

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26 May 2017, 11:10
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55% (hard)

Question Stats:

62% (02:02) correct 38% (02:29) wrong based on 143 sessions

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If p is a positive integer and r is the remainder when p^2-1 is divided by 8, what is the value of r ?

(1) When p is divided by 4, the remainder is zero.
(2) When p is divided by 8, the remainder is not zero.

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Joined: 03 Feb 2017
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Re: If p is a positive integer and r is the remainder when p2 1 is divide  [#permalink]

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01 Jun 2017, 04:39
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Plugging in some numbers for P to check conditions.

A--> P=4, 8, 12 etc all divisible by 4, but (P^2-1) gives reminder as 7 always. SUFFICIENT
B--> If P=2, (P^2-1) reminder is 3. But, if P=3, (P^2-1) reminder is 0. Two different answers. INSUFFICIENT

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Re: If p is a positive integer and r is the remainder when p2 1 is divide  [#permalink]

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22 Aug 2017, 18:01
1
Plugged in numbers to solve the question

Given p>0 and P^2-1/8 = quotient + r

From

1) Using p=4,8,12 etc we see the reminder is constant(7). Sufficient

2) Here p can be 1,2,3,4,5,6,7 but for each value of - the reminder will change. Hence No sufficient

Hence A
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Re: If p is a positive integer and r is the remainder when p2 1 is divide  [#permalink]

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20 Sep 2017, 08:30
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If p is a positive integer and r is the remainder when p^2-1 is divided by 8, what is the value of r ?

(1) When p is divided by 4, the remainder is zero.
(2) When p is divided by 8, the remainder is not zero.

$$p^2-1 = 8q+r$$, or $$p^2 = 8q+r+1$$

Statement 1: implies that $$p=4k$$, so $$p^2 = 16k^2$$
This means $$p^2$$ is divisible by $$8$$. Hence $$r+1=8$$, or $$r=7$$. Sufficient

Statement 2: here the remainder can be anything from 1 to 7. Hence Insufficient

Option A
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Re: If p is a positive integer and r is the remainder when p2 1 is divide  [#permalink]

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21 Sep 2017, 02:28
Hey ! I was able to solve this question by taking values but can you tell how did you get
r+1 = 8 ? niks18
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Re: If p is a positive integer and r is the remainder when p2 1 is divide  [#permalink]

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21 Sep 2017, 02:55
kunalsinghNS wrote:
Hey ! I was able to solve this question by taking values but can you tell how did you get
r+1 = 8 ? niks18

Hi kunalsinghNS

We know that $$p^2$$ is divisible by $$8$$ and

$$p^2 = 8q + (r+1)$$. As $$8q$$ is divisible by $$8$$, so $$(r+1)$$ has to be divisible by $$8$$ to make $$p^2$$ divisible by $$8$$

here $$r$$ is a remainder when $$p^2-1$$ was divided by $$8$$ hence $$r<8$$, so the maximum value $$(r+1)$$ can take is $$8$$

Hence $$r+1=8$$, or $$r=7$$
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Re: If p is a positive integer and r is the remainder when p2 1 is divide  [#permalink]

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21 Sep 2017, 04:39
Thank you niks18
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Re: If p is a positive integer and r is the remainder when p2 1 is divide  [#permalink]

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06 Nov 2017, 23:37
$$(p^2 - 1)$$ % 8 = ?

Statement 1:
Let p = 4m (since it is divisible by 4)
$$(4m - 1)(4m + 1$$) % 8 = ?

$$16m^2 - 1$$ => since $$16m^2$$ is multiple of 8, any value 1 less than multiple of 8, when divided by 8 will yield remainder 7.

Sufficient

Statement 2:
p is even but not divisible by 8, $$(p-1)(p+1$$) =>$$(odd * odd)$$ %8 => can lead to any value between 0 and 7
p is odd , say p = 9, then $$(p-1)(p+1)$$ => (8 * 10) % 8 => remainder 0

InSuff

Re: If p is a positive integer and r is the remainder when p2 1 is divide &nbs [#permalink] 06 Nov 2017, 23:37
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