If r and s are positive integers and r = s + 2, is r + s divisible by

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

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

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

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

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

Answer (A)

From first statement
R-S =2
Where R is a prime number so R is either 6k +1 or 6k-1 form where k is an integer.
Now, if R is 6k+1 form then S is 6k -1 form, so their difference is 2
Therefore R+ S = 12k
Hence divisible by 4
Now, if R is 6k-1 form S is 6k+1 form, so their difference is 2
Therefore R + S =12k
Hence given statement is sufficient
From second statement
S is a prime number
R =3 +2 =5
R + S = 5+3=8
Divisible by 4
R = 2+2 =4
R +S = 4+2 =6
Not divisible by 4
Hence not sufficient
Therefore answer is option A

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