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16 Sep 2014, 00:36
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If \(p\) is a prime number greater than 2, what is the remainder when \(p^2\) is divided by 4?
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
16 Sep 2014, 00:36
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Official Solution:
If \(p\) is a prime number greater than 2, what is the remainder when \(p^2\) is divided by 4?
A. 0
B. 1
C. 2
D. 3
E. 4
There are several algebraic methods to solve this question. However, the easiest way is as follows: since in a PS question two correct answers cannot exist , just choose a prime greater than 2, square it, and observe the remainder upon dividing by 4.
If \(p=3\), then \(p^2=9\), and the remainder upon dividing 9 by 4 is 1.
Nevertheless, if you are interested, here is how to solve the question algebraically. Since it's given that \(p\) is a prime number greater than 2, it must be odd. Therefore, we can express it as \(p = 2k+1\), for some positive integer \(k\). Thus, \(p^2=(2k+1)^2=4k^2+4k+1\). The first two terms, \(4k^2\) and \(4k\), are divisible by 4, which means the remainder will originate from the third term 1, which, when divided by 4, yields a remainder of 1.
Answer: B
If \(p\) is a prime number greater than 2, what is the remainder when \(p^2\) is divided by 4?
A. 0
B. 1
C. 2
D. 3
E. 4
There are several algebraic methods to solve this question. However, the easiest way is as follows: since in a PS question two correct answers cannot exist , just choose a prime greater than 2, square it, and observe the remainder upon dividing by 4.
If \(p=3\), then \(p^2=9\), and the remainder upon dividing 9 by 4 is 1.
Nevertheless, if you are interested, here is how to solve the question algebraically. Since it's given that \(p\) is a prime number greater than 2, it must be odd. Therefore, we can express it as \(p = 2k+1\), for some positive integer \(k\). Thus, \(p^2=(2k+1)^2=4k^2+4k+1\). The first two terms, \(4k^2\) and \(4k\), are divisible by 4, which means the remainder will originate from the third term 1, which, when divided by 4, yields a remainder of 1.
Answer: B
12 Aug 2023, 02:13
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I think this is a high-quality question and I agree with explanation.
