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

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

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10 Sep 2017, 01:22
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85% (hard)

Question Stats:

47% (02:05) correct 53% (02:11) wrong based on 82 sessions

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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

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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, 01:54
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.

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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, 04:46
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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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
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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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Updated on: 10 Sep 2017, 06:58
1
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

Originally posted by niks18 on 10 Sep 2017, 06:55.
Last edited by niks18 on 10 Sep 2017, 06:58, edited 1 time in total.
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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, 06:58
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, is r + s divisible by  [#permalink]

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10 Sep 2017, 07:04
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, 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

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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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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

Re: If r and s are positive integers and r = s + 2, is r + s divisible by &nbs [#permalink] 14 Nov 2017, 12:20
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