I couldn't find this problem with a search. I can't seem to crack this one...

If r is the remainder when the positive integer n is divided by 7, what is the value of r?

(1) When n is divided by 21, the remainder is an odd number

(2) when n is divided by 28, the remainder is 3

Answer is B

===

I tried rephrasing (1) as n/21=2k+1
and (2) as n/28=k+3

I'm able to to manipulate both to get n/7 on the left, but I get different remainders. Not sure where I'm going wrong.

Thanks

..This is the question stem:

Let's take Statement 2 first as it looks simpler:


So what can the the values for n be?
n=28*1+3=31(31/28 leaves a remainder 3),28*2+3=59(51/28 again leaves a remainder 3)...et al..

Substituting n=31 in the question stem, we get,
31/7 leaves a remainder 3. Again, substituting n=59, we get, 59/7, which again leaves a remainder 3. Thus whatever, (multiple of 28)+3 you take and divide by 7, you will consistently get a remainder of 3. Hence B is Sufficient.

Let's consider Statement 1:


n can be 22( 22/21 leaves a remainder 1, which is odd),24(24/21 leaves a remainder 3, which is again odd)....so on and so forth...

Substituting n=22 in the question stem, we get,
22/7 leaves a remainder 1. Substituting n=24;24/7 leaves a remainder 3. Hence the remainder values are inconsistent. Hence A is insufficient

Hope you understood!!

this is the sort of problem that is about 1000 times as hard to solve with theory as it is to solve just by picking numbers. WITH DS PROBLEMS ABOUT REMAINDERS, YOU SHOULD STRONGLY CONSIDER NUMBER-PICKING SOLUTIONS.

watch how simple the number-picking solution is:


let's just pick some numbers that give odd remainders upon division by 21.

* how about n = 22 (gives a remainder of 1 upon division by 21, so satisfies the statement)
--> when you divide this by 7, you get a remainder of 1


* how about n = 24 (gives a remainder of 3 upon division by 21, so satisfies the statement)
--> when you divide this by 7, you get a remainder of 3

insufficient.




let's just pick some numbers that give a remainder of 3 upon division by 28.
to get these, add 3 to a bunch of multiples of 28
--> 31, 59, 87, 115, 143, etc.
if you divide these by 7, you'll notice that the remainders are all 3.
--> clear PATTERN of the same # every time
--> sufficient!

--

by the way, the theory way to solve #2 is:
write n = 28k + 3, where k is an integer
therefore, since 28k is a multiple of 7, that part won't contribute to the remainder when you divide by 7; thus the remainder will be 3.

This makes a lot of sense. Thanks for the quick reply!!

glad to hear it!

_________________
Tim Sanders
Manhattan GMAT Instructor

Ron:

Would statement 1 qualify for an outright rejection (quite like my application to Harvard), because the moment we see "the remainder is an odd number", we must think, this is insufficient-the remainder could be both 1 and 9997797977911.

I mean, is this one of those DS questions that can be solved under a minute and a half?

Well easy way to solve under 30 sec would be

Reminder for n/7 could be always either of 1,2,3,4,5, or 6.

A: n/21 is ODD
which is n/(7*3) = ODD ==> n/7= ODD *3 = EVEN
which is (2,4,6) So INSUFFICIENT

B: n/28 = 3 (Definite value - So this may be the potential answer)
n/(7*2*2) = 3
n/7 = 12 ==> remainder is 5 so it a definite answer

well, no. a remainder must be smaller than the number you are originally dividing by.
so, the remainder produced by dividing an integer by 21 must be somewhere from 0 to 20. also, the answer to the problem itself has to be an integer from 0 to 6.

still, it shouldn't take you very long to find cases that prove this statement insufficient.
for instance, when 22 is divided by 21, the remainder is 1, so n can be 22. if n is 22, then the answer to the question is 1.
when 24 is divided by 21, the remainder is 1, so n can be 24 as well. if n is 24, then the answer to the question is 3.
that's two different answers to the question, so, insufficient.

amazingly, the vast majority of posts on this forum that start with “oh, this is easy” turn out to be wrong. this is no exception.

please be very, very careful about using words like “easy” in a forum that, by definition, is full of struggling learners. not only is it a disrespectful cheap shot against people who find the problem difficult, but it will also come back to bite you if your own solution is incorrect (as is the case here).




this work is 100% incorrect; if it gives you the correct answer to the question, then that is pure luck.

the problem statement is talking about remainders; this work has nothing whatsoever to do with remainders -- you are writing equations that deal with the quotient itself, not the remainder. (i.e., “n/21 is odd” does not mean that the remainder from dividing n by 21 is odd; it means that the quotient itself is odd. in fact, if n/21 is an odd integer, then the remainder upon dividing n by 21 can't be odd, because it will always be 0.)

please read the posts above to get a sense of what you actually can do on this problem; you'll have to start from scratch.

note also that the second statement (the one that is sufficient) actually gives an answer of r = 3, not r = 5.