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If r and s are integers, is r+s divisible by 3?
(1) s is divisible by 3
(2) r is not divisible by 3
(1) s is divisible by 3
(2) r is not divisible by 3
08 Feb 2012, 02:11
Kudos
Bookmarks
If r and s are integers, is r+s divisible by 3?
(1) s is divisible by 3 --> s={a multiple of 3}. Not sufficient as no info about r.
(2) r is not divisible by 3 --> r={NOT a multiple of 3}. Not sufficient as no info about s.
(1)+(2) r+s={NOT multiple of 3}+{a multiple of 3}={NOT 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.
(1) s is divisible by 3 --> s={a multiple of 3}. Not sufficient as no info about r.
(2) r is not divisible by 3 --> r={NOT a multiple of 3}. Not sufficient as no info about s.
(1)+(2) r+s={NOT multiple of 3}+{a multiple of 3}={NOT 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.
07 Feb 2012, 23:29
Kudos
Bookmarks
Hi, there. I'm happy to help with this --- it looks like a cool question. 
Prompt: if r and s are integers, is r+s divisible by 3?
There are two ways to add up two integers and get a sum that's divisible by 3.
(a) both r and s individually are multiples of 3, so their sum will be a multiple of 3. (Example: 9 + 6 = 15)
(b) let's say that neither r nor s is divisible by 3; let's say that when we divide r by 3, the remainder is m, and when we divide s by 3, the remainder is n; if m + n = 3, then r + s will be a multiple of 3.
Example:
When 5 is divided by 3, remainder = 2
When 7 is divided by 3, remainder = 1
2+1 = 3 ----> 5 + 7 = 12
The basic idea there is that the sum of the remainders will be the remainder of the sum. If sum of the remainders is divisible by 3, then when the sum of the numbers is divided by three, it will go into the sum of the remainders evenly, and therefore go into the entire sum evenly.
BTW, in everything I've said between the prompt and this line, you could replace 3 with any other positive integer greater than 1, and it would still be true.
If we're in one of these two possibilities, then we will know that r+s is divisible by 3.
Statement #1: s is divisible by 3
Promising, but we know nothing about r, so this statement, by itself, is insufficient.
Statement #2: r is not divisible by 3
Again, this could be a step in the right direction, but now we know nothing about s, so this statement, by itself, is insufficient.
Combined Statements #1 & #2: s is divisible by 3 and r is not divisible by 3
Well, neither of the scenarios above allow for one number divisible by 3 and the other not divisible by 3. In fact, when we divide (r+s) by 3, we know the s part will not have a remainder but the r part will, which means that 3 does not go evenly into the sum r+s. Therefore, we can give a definitive "no" answer to the prompt question, and because we have the ability to give a definitive answer, that means we must have sufficient information.
Statements are insufficient individually but are sufficient combined ==> Answer = C.
Does all that make sense?
Here's a particularly difficult divisibility DS question, for more practice.
https://gmat.magoosh.com/questions/871
Let me know if you have any questions about anything I have said here.
Mike
Prompt: if r and s are integers, is r+s divisible by 3?
There are two ways to add up two integers and get a sum that's divisible by 3.
(a) both r and s individually are multiples of 3, so their sum will be a multiple of 3. (Example: 9 + 6 = 15)
(b) let's say that neither r nor s is divisible by 3; let's say that when we divide r by 3, the remainder is m, and when we divide s by 3, the remainder is n; if m + n = 3, then r + s will be a multiple of 3.
Example:
When 5 is divided by 3, remainder = 2
When 7 is divided by 3, remainder = 1
2+1 = 3 ----> 5 + 7 = 12
The basic idea there is that the sum of the remainders will be the remainder of the sum. If sum of the remainders is divisible by 3, then when the sum of the numbers is divided by three, it will go into the sum of the remainders evenly, and therefore go into the entire sum evenly.
BTW, in everything I've said between the prompt and this line, you could replace 3 with any other positive integer greater than 1, and it would still be true.
If we're in one of these two possibilities, then we will know that r+s is divisible by 3.
Statement #1: s is divisible by 3
Promising, but we know nothing about r, so this statement, by itself, is insufficient.
Statement #2: r is not divisible by 3
Again, this could be a step in the right direction, but now we know nothing about s, so this statement, by itself, is insufficient.
Combined Statements #1 & #2: s is divisible by 3 and r is not divisible by 3
Well, neither of the scenarios above allow for one number divisible by 3 and the other not divisible by 3. In fact, when we divide (r+s) by 3, we know the s part will not have a remainder but the r part will, which means that 3 does not go evenly into the sum r+s. Therefore, we can give a definitive "no" answer to the prompt question, and because we have the ability to give a definitive answer, that means we must have sufficient information.
Statements are insufficient individually but are sufficient combined ==> Answer = C.
Does all that make sense?
Here's a particularly difficult divisibility DS question, for more practice.
https://gmat.magoosh.com/questions/871
Let me know if you have any questions about anything I have said here.
Mike
