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

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10 Sep 2017, 06:07

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 Please tell OA, my exam is on 15th of this month

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

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10 Sep 2017, 06:55

1

This post received KUDOS

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, 06:58, edited 1 time in total.

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

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10 Sep 2017, 07:33

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

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

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14 Nov 2017, 12:20

\(r = s + 2\) => \(r - s =2\)

Statement 1: \(r\) is prime number, \(r\) can't be 2, if it were, then \(s = 0\) but \(s\) is positive so \(r\) is odd prime number, any odd number can be written as \((2p + 1)\) or \((2k + 3)\) or \((2l + 5)\).. where \(p,k,l\)are integers Let \(r = 2k + 3\), then \(s = r - 2\) => \(2k + 1\) so \(r + s = (2k + 3 + 2k + 1) = (4k + 4)\) => which is clearly divisible by 4 => suff

Statement 2: \(s\) is prime number. if \(s = 2 => r = 4 => (r+s) = 6\) not divisible by 4 if \(s\) is odd prime number, we can use above method in statement 1, to prove that it is divisible by 4 so two diff possiblities => Not Sufficient