kunalbh19 wrote:
Bunuel wrote:
enigma123 wrote:
If r – s = 3p, is p an integer?
(1) r is divisible by 735
(2) r + s is divisible by 3
If r – s = 3p , is p an integer?Question basically asks whether r-s is a multiple of 3 (..., -6, -3, 0, 3, 6, ...), because if it is then p would be an integer.
(1) r is divisible by 735 --> r is a multiple of 3 (as the sum of the digits of 735 is divisible by 3), though this statement is insufficient as no info about s.
(2) r + s is divisible by 3 --> r + s is a multiple of 3. Now, if r=2 and s=1 then r-s=1 and the answer is NO but if r=s=0 then r-s=0 and the answer is YES. Not sufficient.
(1)+(2) From (1) r={multiple of 3} then as from (2) r+s={multiple of 3}+s={multiple of 3} then s is also a multiple of 3 --> r-s={multiple of 3}-{multiple of 3}={multiple of 3}. Sufficient.
Answer: C.
Below might help to understand this concept better.
If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.
If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.
If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3;
OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5;
OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.
Hope it's clear.
Hi,
I think B should be sufficient. Here is how:
We are given r+s is divisible by 3.
r+s=3a (where a is any integer) -----(1)
r-s+2s = 3a
3p + 2s = 3a (since we are given that r-s=3p)
2s = 3(a-p)
since 2 and 3 are both primes, we can conclude that a-p is a multiple of 2 and s is a multiple of 3.
similarly if we replace r by (-r+2r), we will get that r is also divisible by 3.
hence, if both r and s are divisible by 3, therefore p is an integer.
please correct me if i am wrong somewhere.
Thanks in advance.
I did the same mistake as you and assumed statement 2 by itself would suffice .
After pondering on it for a while , figured where I went wrong.
To answer your question , consider your below explanation -
r+s=3a (where a is any integer) -----(1)
r-s+2s = 3a
3p + 2s = 3a (since we are given that r-s=3p)
2s = 3(a-p)
since 2 and 3 are both primes, we can conclude that a-p is a multiple of 2 and s is a multiple of 3.Let me try to explain why statement 2 can be insufficient.
2s=3(a-p) . you got it right till here.
But you cannot conclude with just statement 2 , that a-p is a multiple of 2 . Cos all we know untill this point is "a" is an integer. We do not know weather S or/and P is an integer .
To illustrate what I mean above, lemme give you an example -
consider 2s=3(a-p) ===> 2 (0.3) = 3(0.2) , where s=0.3 and a-p=0.2 , a is an integer , lets say 2 , in which case P would be 1.8 . Hence we can prove that p is not an integer.
similarly , we can prove otherwise that P is an integer.
Now if you consider statement 1 , which says S is an integer , we can conclude from the equation 2S=3(a-p) , that P is an integer.
cos 2*integer = 3 * integer , i.e a-p SHOULD be an integer , since a is an integer and S is an integer .
Hope that helps
Jyothi