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].
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
