n = 7*q + k, with r being between 1 and 6 inclusively.
n^2 = (7q + k)(7q + k) = 49q^2 + 14qk + k^2.
As you may notice, 49q^2 + 14qk is divisible by 7, leaving the question of a remainder to k^2.
Now, k^2 might be smaller than 7 or greater than 7. This is why we get two cases:
a. k^2 is smaller than 7, meaning that k is either 1 or 2. In this case, you get that the remainder of n^2 divided by 7 is exactly k^2.
Now, for k = 1, k^2 = 1 - so this is a good one. You could stop here, and this is advised during the exam. Now, we will continue jsut for the sake of exercise... For k = 2, then k^2 will not be the same as k, so 2 is out.

b. k^2 is greater than 7, meaning that the remainder of dividing n^2 by 7 will not be k^2, but rather the remainder of dividing k^2. Here you get 4 values for k: 3, 4, 5 and 6. But since we get only 4 and 6 in the answers, we're just going to use these and not waste time with 3 and 5.
k = 4 makes k^2 = 16. The remainder of 16 : 7 is 2, which is not equal to 4. So k is out.
k = 6 makes k^2 = 36, with remainder 1 (since 35 = 5*7), which is again not equal to 6.

sanju09 wrote:The positive integer n is not divisible by 7. The remainder when n^2 is divided by 7 and the remainder when n is divided by 7 are each equal to k. What is k?

(A) 1
(B) 2
(C) 4
(D) 6
(E) It cannot be determined from the information given.

I plunge the values from the option and find A is the correct answer.

Thanks,
Farooq.

sanju09 wrote:The positive integer n is not divisible by 7. The remainder when n^2 is divided by 7 and the remainder when n is divided by 7 are each equal to k. What is k?

(A) 1
(B) 2
(C) 4
(D) 6
(E) It cannot be determined from the information given.

Trial and error

let n=8

8/7 = 1 R 1
8^2 = 64 / 7 = 9 R 1

Both have reminder 1, we have satisfied our condition no need to evaluate any other answer.

A.

Please start posting the OA

Since the positive integer n leaves nonzero remainder k when divided by 7, it can be written as n = 7a + k, where a is a nonnegative integer and k is equal to one of the values 1, 2, 3, 4, 5, or 6. Since n^2 leaves the same nonzero remainder k when divided by 7, it can be written as n = 7b + k, where b is a nonnegative integer and k has the same value. It is also true that n^2 = (7a + k)^2 = 49a^2 + 14ak + k^2, which can be written as 7(7a^2 + 2ak) + k^2. Since n^2 = 7b + k = 7(7a^2 + 2ak) + k^2, it follows that k and k^2 leave the same remainder when divided by 7.

We know that k is equal to one of the values 1, 2, 3, 4, 5, or 6. We can see which of these values for k has the property that k and k^2 leave the same remainder when divided by 7. If k = 1, then k^2 = 1, which leaves remainder 1 when divided by 7. Thus, 1 is a possible value for k. But we are not done yet; since one of the answer choices is (E), It cannot be determined from the information given, we must continue and check 2, 3, 4, 5, and 6 as possible values for k.

If k = 2, then k^2 = 4, which leaves remainder 4 ≠ 2 when divided by 7. If k = 3, then k^2 = 9, which leaves remainder 2 ≠ 3 when divided by 7. If k = 4, then k^2 = 16, which leaves remainder 2 ≠ 4 when divided by 7. If k = 5, then k^2 = 25, which leaves remainder 4 ≠ 5 when divided by 7. If k = 6, then k^2 = 36, which leaves remainder 1 ≠ 6 when divided by 7.

Therefore, the only possible value of k is 1, which is choice [spoiler](A)[/spoiler].