Say, \(n^2 - 8\) gives remainder \(R\) when divided by 33.
for simplicity let us take only \(n^2\) thus, \(n^2\) gives remainder \(8 + R\) when divided by 33.
However, 33 has factors 3 and 11, \(n^2\) must also satisfy congruency when divided by 3 and 11 also.
Now take 3 since it is smaller among the 2 factors of 33.
when \(n = 0\), \(n^2 = 0\) , remainder when divided by 3 = 0
when \(n = 1\), \(n^2 = 1\) , remainder when divided by 3 = 1
when \(n = 2\), \(n^2 = 4\) , remainder when divided by 3 = 1
Thus, we will find that \(n^2\) for any positive integer n , remainder when divided by 3 will be either 0 or 1
Now test the 3 cases in the answer choices.
I. For \(R = 0\), \(n^2\) gives remainder \(8 + R = 8\) when divided by 33.
Now let us check what remainder we will get when divided by 3, by further dividing 8 by 3 we get remainder 2. but we should get either 0 or 1. Hence \(R = 0\) is not possible.
II. For \(R = 15\), similarly \(8 + R = 23\), when divided by 3 gives remainder 2. Hence \(R = 15\) also not possible.
III. For \(R = 21\), again \(8 + R = 29\), when divided by 3 gives remainder 2. Hence \(R = 21\) also not possible.
Answer is E. None of these
Bunuel
If n is a positive integer, which of the following could be the remainder when n^2 - 8 is divided by 33?
I. 0
II. 15
III. 21
A. I only
B. II only
C. III only
D. II and III only
E. None of these
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