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If r and s are positive integers and r=s+2

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If r and s are positive integers and r=s+2 [#permalink]

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If r and s are positive integers and r=s+2, is r+s divisible by 4?
(1) r is a prime number.
(2) s is a prime number.


A new tricky DS Question
[Reveal] Spoiler: OA

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Re: If r and s are positive integers and r=s+2 [#permalink]

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New post 10 Sep 2017, 02:54
chetan2u wrote:
If r and s are positive integers and r=s+2, is r+s divisible by 4?
(1) r is a prime number.
(2) s is a prime number.


A new tricky DS Question


If r and s are both even, then r+s isn't divisible by 4.
If r and s are both odd, then r+s is divisible by 4.

(1) r is a prime number.

r = s+2 so r > 2. Hence r must be odd, so s must be odd.

Thus, r+s is divisible by 4. Sufficient.

(2) s is a prime number.

If s=2, then s is even and r is even. Thus r+s isn't divisible by 4.
If s=3, then s is odd and r is odd. Thus r+s is divisible by 4.

Insufficient.

Answer A.
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Re: If r and s are positive integers and r=s+2 [#permalink]

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New post 10 Sep 2017, 05:46
chetan2u wrote:
If r and s are positive integers and r=s+2, is r+s divisible by 4?
(1) r is a prime number.
(2) s is a prime number.


A new tricky DS Question


Ans is B.
r+s = 2(s+1) via substitution of r= s+2

if s is odd/prime, then r+2 should be divisible

On the other hand, 2(r-1) [again substitute] will not work with if r is 2, work if r =3

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If r and s are positive integers and r=s+2 [#permalink]

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New post 10 Sep 2017, 07:07
A??
r+s = 2(s+1)
st1> r is prime, so s is also odd as r cannot be 2(if r=2 then s=0). Other than 2 for every prime R, S will be odd. Odd + 1 is even so S+1 is even.
Even * 2 will be a multiple of 4.
suff
St2> If s is prime, S+1 will be even,
Even * 2 will be a multiple of 4.

But if S=2, then r=4.
in-suff

hence A
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If r and s are positive integers and r=s+2 [#permalink]

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New post 10 Sep 2017, 07:55
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chetan2u wrote:
If r and s are positive integers and r=s+2, is r+s divisible by 4?
(1) r is a prime number.
(2) s is a prime number.


A new tricky DS Question

\(r+s = r+r-2\) (as \(r=s+2\), so \(s=r-2\))
This implies: \(2r-2=2(r-1)\).
hence \(\frac{(r+s)}{4} = \frac{2(r-1)}{4} = \frac{(r-1)}{2}\)-------------\((1)\)
So for \((r+s)\) to be divisible by \(4\), \(r\) needs to be ODD

Statement 1: this implies that \(r\) is ODD. Hence sufficient
(note: \(r= s+2\) and \(s\) is a positive integer, hence \(s\) cannot be \(0\), so \(r\) cannot be \(2\), the only even prime no)

Statement 2: if \(s =2\), then \(r=4 = Even\), hence equation \((1)\) is not divisible by \(4\)
but if \(s =\) odd prime no for eg. \(3\), then \(r = 5 = Odd\), hence equation \((1)\) is divisible by \(4\). Hence the statement is Insufficient

Option A

Last edited by niks18 on 10 Sep 2017, 07:58, edited 1 time in total.

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Re: If r and s are positive integers and r=s+2 [#permalink]

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New post 10 Sep 2017, 07:58
B.
r + s : 4 is equivalent to (r + s) : 4 =(s+s+2):4=2*(s+2):4=(s+2):2. Hence, s must be an even number for residual to be equal to 0.

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Re: If r and s are positive integers and r=s+2 [#permalink]

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New post 10 Sep 2017, 08:04
AlexGmat2017 wrote:
B.
r + s : 4 is equivalent to (r + s) : 4 =(s+s+2):4=2*(s+2):4=(s+2):2. Hence, s must be an even number for residual to be equal to 0.


Hi AlexGmat2017

as per statement 2 "s" is prime no so it can be even for eg. 2 or it can be odd for eg. 3. the highlighted section is not always true

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Re: If r and s are positive integers and r=s+2 [#permalink]

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New post 10 Sep 2017, 08:33
A . R=s+2, statement 1 - r is prime which means s can take values of 1,3,5,9 etc so thay r would be 3,5,7,11 and so on and adding r+s would be 4,8,12,20,24 which are all multiples of 4 and hence is sufficient.
Statement 2- s is prime means s can take values of 2,3,5,7,11 etc and r cam take values of 4,5,7,9,13 and r+s takes values of 6,8,12,16,24 and so on and 6 is not divisible by 4 hence 8,12,16,24 are, hence insufficient. So answer A

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Re: If r and s are positive integers and r=s+2   [#permalink] 10 Sep 2017, 08:33
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