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It is given that p, q, and r are positive integers, where p is an odd

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It is given that p, q, and r are positive integers, where p is an odd  [#permalink]

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New post 13 Oct 2018, 09:04
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It is given that p, q, and r are positive integers, where p is an odd number and \(r = p^2 + q^2 + 4.\) Is \(q^3\) divisible by 8?
(1) r = 8k -3 where k is a positive integer.
(2) When (r-p+1) is divided by 2, it leaves a remainder.

Weekly Quant Quiz #4 Ques 2


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Re: It is given that p, q, and r are positive integers, where p is an odd  [#permalink]

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New post 13 Oct 2018, 09:11
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According to me (D) should be the answer.

Please find a solution attached.

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Re: It is given that p, q, and r are positive integers, where p is an odd  [#permalink]

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New post 13 Oct 2018, 09:16
From question stem we can derive that q will be odd if r is even and q will be even if r is odd.

From Statement - 1:
r = 8k-3 basically means r is always a positive odd integer. This implies q can only be even which means q has 2 as its factor and so we can say q3 is divisible by 8.

From statement - 2:
Because p is odd, and (r-p+1)/2 is also odd we can conclude that r is odd.
This implies q can only be even which means q has 2 as its factor and so we can say q3 is divisible by 8.
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Re: It is given that p, q, and r are positive integers, where p is an odd  [#permalink]

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New post 13 Oct 2018, 09:27
given p is odd , and r = p^2+q^2+4

statement : 1 ==>r = 8k -3 where k is a positive integer

it does not give any information about Q , so insuffient .


Statement 2 :When (r-p+1) is divided by 2, it leaves a remainder.
r-p+1 is not divisible by 2 , so it says that r-p+1 is a odd number
r-p+1 = Odd
==> r-p = Odd+1 = Even
==> given p is Odd
==> r = Even + p(Odd)
==> r = Even +Odd = Odd
So now we have P ( Odd ) , r(Odd)
so r = Odd is never divisible by 8 . Sufficient , So OA is B
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Re: It is given that p, q, and r are positive integers, where p is an odd  [#permalink]

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New post 13 Oct 2018, 09:27
It is given that p, q, and r are positive integers, where p is an odd number and r=p2+q2+4.r=p2+q2+4. Is q3q3 divisible by 8?
(1) r = 8k -3 where k is a positive integer.
(2) When (r-p+1) is divided by 2, it leaves a remainder.

from given eqn 2


no value of r given
q2=r-p2-4 ( is a factor of 8 or not)
From stmnt1: for value of k=1 we get r= 5 and k=2 r= 13

plugging into eqn we get yes and no insufficient

from stmtn2

IMO C
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Re: It is given that p, q, and r are positive integers, where p is an odd  [#permalink]

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New post 13 Oct 2018, 09:29
r=p^2+q^2+4 p odd p^2 odd p^2 +4 odd

(1) r = 8k -3 where k is a positive int------ Suff
r-3 even, r odd
p^2+q^2+1 even , p^2+q^2 odd, q^2 even , q even , q=2a hence q^3 = 8a^3


(2) When (r-p+1) is divided by 2, it leaves a remainder. - Suff
r-p even , r odd ( as p odd) ====>similar to 1==>q^3 = 8a^3

Ans D
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Re: It is given that p, q, and r are positive integers, where p is an odd  [#permalink]

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New post 13 Oct 2018, 11:28
The answer is fine.But how the hell do we solve this question in under 2 mins..?

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It is given that p, q, and r are positive integers, where p is an odd  [#permalink]

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New post 13 Oct 2018, 19:30
Hi redskull1
Please see the explanations above, you will understand the solution.
If you still have any doubt, feel free to tag me :angel:

redskull1 wrote:
The answer is fine.But how the hell do we solve this question in under 2 mins..?

Posted from my mobile device

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It is given that p, q, and r are positive integers, where p is an odd   [#permalink] 13 Oct 2018, 19:30
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