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21 Feb 2020, 05:52
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B
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Difficulty:
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Question Stats:
59% (01:51) correct
41%
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For how many prime numbers p, is p^2 + 3p - 1 a prime number?
A. 0
B. 1
C. 2
D. 3
E. 4
Are You Up For the Challenge: 700 Level Questions
A. 0
B. 1
C. 2
D. 3
E. 4
Are You Up For the Challenge: 700 Level Questions
21 Feb 2020, 10:23
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Any prime number greater than 3 is in the form of 6k±1
1. when p=6k+1, where k is an integer
\( p^2 + 3p - 1= 36k^2+12k+1+18k+3-1= 36k^2+30k+3= 3(12k^2+10k+1) \)
hence, it will always a multiple of 3
2. when p=6k-1, where k is an integer
\(p^2 + 3p - 1= 36k^2-12k+1+18k-3-1= 36k^2+6k-3= 3(12k^2+2k-1) \)
hence, it will always a multiple of 3
\( p^2 + 3p - 1\) can never be a prime number, if p is a prime number greater than 3
Now we can check for p=2 and p=3
when p=2, \(p^2 + 3p - 1\)= 9 (not a prime number)
when p=3,\( p^2 + 3p - \)1= 17 (prime number)
B
1. when p=6k+1, where k is an integer
\( p^2 + 3p - 1= 36k^2+12k+1+18k+3-1= 36k^2+30k+3= 3(12k^2+10k+1) \)
hence, it will always a multiple of 3
2. when p=6k-1, where k is an integer
\(p^2 + 3p - 1= 36k^2-12k+1+18k-3-1= 36k^2+6k-3= 3(12k^2+2k-1) \)
hence, it will always a multiple of 3
\( p^2 + 3p - 1\) can never be a prime number, if p is a prime number greater than 3
Now we can check for p=2 and p=3
when p=2, \(p^2 + 3p - 1\)= 9 (not a prime number)
when p=3,\( p^2 + 3p - \)1= 17 (prime number)
B
Bunuel
General Discussion
21 Feb 2020, 06:00
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Bunuel
@p = 2, p^2 + 3p - 1=2+6-1 = 9 NOT Prime
@p = 3, p^2 + 3p - 1=9+9-1 = 17 Prime
@p = 5, p^2 + 3p - 1=25+15-1 = 39 NOT Prime
@p = 7, p^2 + 3p - 1=49+21-1 = 69 NOT Prime
@p = 11, p^2 + 3p - 1=121+33-1 = 153 NOT Prime
For any prime value of p other than 3, the value of expression p^2 + 3p - 1 is always divisible by 3 hence there will NOT be any next value of p that satisfies the given constraint
Answer: Option B
