What is integer n? 1. When divided by 7, remainder is 3 2.

Yes it is E....

I find another set of values

when y =9 and x=5 in my solution but your looks more elegant, laxieqv

Thanks!

Laxieqv ... thats super clever .. i havent seen that before but can you explain where this theorem came from ? and where else it could be used ?

you can check it out here https://www.gmatclub.com/phpbb/viewtopic.php?t=27093

E.

n = 36,66....it seems strange but I found these values under 2 .

I am not sure on the GMAT I will figure out that "C" is the trap....unless it says that the question is from vivek .

LAXI can u explain how you got this
---> 4v= 1( mod 7) ---> v= 2 (mod 7)

The OA is "E"

Laxie, thanks for the logic buddy!

That is very elegant Laxie!

For people who are a little weary of the whole sequence of the Chinese modulus theorem, I hope this will help: You don't have to memorize it.

Basically you write the two conditions:
n=7a+3
n=4b+2

Therefore 4b=7a-1=8a-a-1
b=2a-(a-1)/4
We know that b is an integer so (a-1)/4 must be an integer.
We can write as a-1=4s, or a=4s+1.

Now we know
a=4s+1,
b=2(4s+1)-s=7s+2

You can solve for n here from either a or b, although you don't have to. As long as you know that when s takes different values, you'll have different values for a and b and thus different values of n. In other words you'll know that the solution is not unique and the answer would be E.

I'm just so impressed by you, sis

The method is neat! And I must say it's the first time I've heard of it. But for people who are stuck, working out with sample numbers present a very quick approach as well.

1) n = 7q1 + 3

n could be 10
n could be 15
etc

Insufficient.

2) n = 4q2 + 1
n could be 5
n could be 9
etc

Insufficient.

Using 1) and 2)
n = 7q1 + 3
n = 4q2 + 1

7q1+3 = 4q2+1
2 = 4q2-7q1

Possible sets: (q1,q2) = (4,2) (11,6) --> means many possible values of n

Ans E

I'm impressed by you all
Thanks Hong, Laxie & Wilfred!

Extension to HongHu's method:

N = 7a + 3
N = 4b + 2

=> 7a+1 is divisible by 4.
Smallest possible positive value of a satisfying above equation is a = 1.
when a = 1, N = 10

LCM of 7,4 = 28

Hence, the set of all such numbers can be written as 28k + 10.

We can extend this procedure to solve more than 2 equations also.