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07 Aug 2026, 03:41
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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
I. 0
II. 15
III. 21
A. I only
B. II only
C. III only
D. II and III only
E. None of these
07 Aug 2026, 03:41
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Official Solution:
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
Since 33 is divisible by 3, first check what remainder \(n^2 - 8\) can leave when divided by 3.
When \(n\) is divided by 3, the remainder can be 0, 1, or 2.
• If \(n\) leaves a remainder of 0, then \(n^2 - 8\) leaves a remainder of 1 when divided by 3.
• If \(n\) leaves a remainder of 1, then \(n^2 - 8 = (3k + 1)^2 - 8 = 9k^2 + 6k - 7\) leaves a remainder of 2 when divided by 3.
• If \(n\) leaves a remainder of 2, then \(n^2 - 8 = (3k + 2)^2 - 8 = 9k^2 + 12k - 4\) leaves a remainder of 2 when divided by 3.
So \(n^2 - 8\) can leave only a remainder of 1 or 2 when divided by 3.
Thus, \(n^2 - 8\) can never be divisible by 3.
But 0, 15, and 21 are all divisible by 3. So if the remainder were any of those, we would get:
\(n^2 - 8 = 33q + r = \{\text{a multiple of 3}\} + \{\text{a multiple of 3}\} = \{\text{a multiple of 3}\}\)
Since \(n^2 - 8\) cannot be a multiple of 3, the remainder cannot be any of 0, 15, or 21.
Answer: E
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
Since 33 is divisible by 3, first check what remainder \(n^2 - 8\) can leave when divided by 3.
When \(n\) is divided by 3, the remainder can be 0, 1, or 2.
• If \(n\) leaves a remainder of 0, then \(n^2 - 8\) leaves a remainder of 1 when divided by 3.
• If \(n\) leaves a remainder of 1, then \(n^2 - 8 = (3k + 1)^2 - 8 = 9k^2 + 6k - 7\) leaves a remainder of 2 when divided by 3.
• If \(n\) leaves a remainder of 2, then \(n^2 - 8 = (3k + 2)^2 - 8 = 9k^2 + 12k - 4\) leaves a remainder of 2 when divided by 3.
So \(n^2 - 8\) can leave only a remainder of 1 or 2 when divided by 3.
Thus, \(n^2 - 8\) can never be divisible by 3.
But 0, 15, and 21 are all divisible by 3. So if the remainder were any of those, we would get:
\(n^2 - 8 = 33q + r = \{\text{a multiple of 3}\} + \{\text{a multiple of 3}\} = \{\text{a multiple of 3}\}\)
Since \(n^2 - 8\) cannot be a multiple of 3, the remainder cannot be any of 0, 15, or 21.
Answer: E
