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16 Sep 2014, 01:27
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If \(p\) is a positive integer, what is the remainder when \(p^2\) is divided by 12?
(1) \(p\) is greater than 3.
(2) \(p\) is a prime.
(1) \(p\) is greater than 3.
(2) \(p\) is a prime.
27 Apr 2016, 00:54
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Official Solution:
If \(p\) is a positive integer, what is the remainder when \(p^2\) is divided by 12?
(1) \(p\) is greater than 3.
This is clearly insufficient as different values of \(p\) yield different remainders.
(2) \(p\) is a prime.
This too is insufficient. For instance, if \(p=2\), the remainder is 4, whereas if \(p=3\), the remainder is 9.
(1)+(2) By plugging in several prime numbers greater than 3, one can observe that the remainder is consistently 1 (for example, using \(p=5\), \(p=7\), or \(p=11\)).
To further verify this using algebra, consider the property of prime numbers: any prime number greater than 3 can be represented as either \(p=6n+1\) or \(p=6n-1\).
For \(p=6n+1\), \(p^2\) becomes \(36n^2+12n+1\), which gives a remainder of 1 when divided by 12.
For \(p=6n-1\), \(p^2\) is \(36n^2-12n+1\), which also results in a remainder of 1 upon division by 12.
Answer: C
If \(p\) is a positive integer, what is the remainder when \(p^2\) is divided by 12?
(1) \(p\) is greater than 3.
This is clearly insufficient as different values of \(p\) yield different remainders.
(2) \(p\) is a prime.
This too is insufficient. For instance, if \(p=2\), the remainder is 4, whereas if \(p=3\), the remainder is 9.
(1)+(2) By plugging in several prime numbers greater than 3, one can observe that the remainder is consistently 1 (for example, using \(p=5\), \(p=7\), or \(p=11\)).
To further verify this using algebra, consider the property of prime numbers: any prime number greater than 3 can be represented as either \(p=6n+1\) or \(p=6n-1\).
For \(p=6n+1\), \(p^2\) becomes \(36n^2+12n+1\), which gives a remainder of 1 when divided by 12.
For \(p=6n-1\), \(p^2\) is \(36n^2-12n+1\), which also results in a remainder of 1 upon division by 12.
Answer: C
