Hi divyalr
This question is a tough one.

After spending some time thinking about it, I believe the answer to this question is the following:
If X is divided by 5, and you get P with a remainder of 3, it means that P is the same result you would get if X-3 was divided by 5 (though this time the remainder would be 0).

Likewise, if X is divided by 11, and you get a certain result (let's say "Z") with a remainder of 3, it means that Z is that the same result you would get if X-3 was divided by 11 (with remainder = 0).
Therefore, it means that X-3 is a multiple of both 5 and 11.

Now, there are two ways of solving this:

One way is to pick numbers: find a number multiple of 5 and 11, divide it by 5 and then by 11, and check if it's an integer or if you have a remainder.
For example: 55 or 110. If you divide 110 by 5 you will get 55, so when you divide 55 by 11, you will get an integer (5), with remainder 0.

The second way requires some knowledge of number properties (after all, this is what this exercise is all about �), tough I am not sure about this one: it seems that if you have a number (Z) which is a multiple of two numbers (say X and Y), the result of the division of Z by X is also a multiple of Y. Or the other way around, the result of the division of Z by Y is always a multiple or X, which is the same to say that Z/X/Y is always an integer.

I know that my answer can be a bit confusing, but I couldn't find this question very clear myself. Please let me know if you have any questions.

divyalr wrote:f positive integer x is divided by 5, the result is p and the remainder 3. If x is divided by 11, the remainder is 3 again, what is the remainder when p is divided by 11?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
Answer is A

Here is how I solved though I feel its very time consuming

Given : x/5=5p+3 -- (1), x/11= 11n+3 --(2)
possible values of x from (1) : 3,8,13,18,23,28...
possible values of x from (1) : 3,14,25,36,47...

therefore x= 3, p = 0, p/11=0

I am wondering if there is a easier method to do the same? Please add your inputs.

Here is what I think:

X = 5p+3

X = 11C+3 (c is any positive integer)

5p + 3 = 11C + 3

Solving we get, 5p = 11 C

P must be 11 & C must be 5. Based on this, remainder will be 0.

HTH.

Based on how crackgmat007 solved the question, with the remainder being 0.
Is the answer to the question "0"?

I would just backsolve and use the answer choices to help you.

First thing that came to mind to have same remainders was to have a number less than the divisor. For example, divisors of 3 and 5 will have a common remainder of 2 from 2/3 and 2/5. This implies that x is most likely 3.

If p = 0, then x = p*5+3. Then that means (0*5+3)/5 will yield R of 3. (0*5+3)/11 will yield R of 3. Voila, first choice works. You can test with other answer choices and see that they don't yield the same remainder.

Or you could do one of those algebraic ways that others described.