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

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New post 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

Answer --> 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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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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SajjadAhmad wrote:
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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New post 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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New post 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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New post 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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New post 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

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

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