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16 Sep 2014, 01:50
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D
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Difficulty:
95%
(hard)
Question Stats:
34% (02:20) correct
66%
(02:21)
wrong
based on 178
sessions
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Time
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Not Attempted Yet
A "Sophie Germain" prime is any positive prime number \(p\) for which \(2p + 1\) is also prime. The product of all the possible units digits of Sophie Germain primes greater than 5 is
A. 3
B. 7
C. 21
D. 27
E. 189
A. 3
B. 7
C. 21
D. 27
E. 189
16 Sep 2014, 01:50
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Official Solution:
A "Sophie Germain" prime is any positive prime number \(p\) for which \(2p + 1\) is also prime. The product of all the possible units digits of Sophie Germain primes greater than 5 is
A. 3
B. 7
C. 21
D. 27
E. 189
The first step is to make sure that we truly understand the definition of these special, so-called "Sophie Germain" primes. The idea is that if both \(p\) and \(2p + 1\) are prime, then \(p\) is a Sophie Germain prime. We can quickly test some small primes to see whether they are Sophie Germain primes:
If \(p = 2\), then \(2p + 1 = 5\), which is also prime. So 2 is a Sophie Germain prime.
If \(p = 3\), then \(2p + 1 = 7\), which is also prime. So 3 is an SG prime.
If \(p = 5\), then \(2p + 1 = 11\), which is also prime. So 5 is an SG prime.
If \(p = 7\), then \(2p + 1 = 15\), which is NOT prime. So 7 is NOT an SG prime.
We are looking for the product of all the possible units digits of SG primes greater than 5.
Let's first narrow down the possible units digits of ALL primes greater than 5. First of all, no prime greater than 2 can end in an even digit (0, 2, 4, 6, 8), since no even number greater than 2 is prime. Likewise, no prime greater than 5 can end in 5, since all integers with a units digit of 5 are divisible by 5. So we can see that primes greater than 5 must end in 1, 3, 7, or 9 (e.g., 11, 13, 17, and 19). The question is whether the SG condition further restricts the possible units digits.
If a prime \(p\) ends in 1, then \(2p + 1\) ends in 3, which is a safe units digit for a prime.
If a prime \(p\) ends in 3, then \(2p + 1\) ends in 7, which is also a safe units digit for a prime.
If a prime \(p\) ends in 7, then \(2p + 1\) ends in 5, which is NOT a safe units digit for a prime greater than 5. So, in fact, no SG prime greater than 5 can end in 7.
Finally, if a prime \(p\) ends in 9, then \(2p + 1\) also ends in 9, which is a safe units digit for a prime.
Thus, SG primes can only end in 1, 3, or 9.
To be rigorous, we should find at least one example of an actual SG prime with each of those units digits, to confirm that we haven't missed something.
If \(p = 11\), then \(2p + 1 = 23\), which is also prime. So 11 is an SG prime, and we have 1 confirmed as a possible SG units digit.
If \(p = 13\), then \(2p + 1 = 27\), which is NOT prime. So 13 is NOT an SG prime.
Let's try 23:
If \(p = 23\), then \(2p + 1 = 47\), which is also prime. So 23 is an SG prime, and we have 3 confirmed as a possible SG units digit.
Finally, let's check 9 as a units digit.
If \(p = 19\), then \(2p + 1 = 39\), which is NOT prime. So 19 is NOT an SG prime.
But if \(p = 29\), then \(2p + 1 = 59\), which IS prime. So 29 is an SG prime, and we have 9 confirmed as a possible SG units digit.
The product of 1, 3, and 9 is 27.
Answer: D
A "Sophie Germain" prime is any positive prime number \(p\) for which \(2p + 1\) is also prime. The product of all the possible units digits of Sophie Germain primes greater than 5 is
A. 3
B. 7
C. 21
D. 27
E. 189
The first step is to make sure that we truly understand the definition of these special, so-called "Sophie Germain" primes. The idea is that if both \(p\) and \(2p + 1\) are prime, then \(p\) is a Sophie Germain prime. We can quickly test some small primes to see whether they are Sophie Germain primes:
If \(p = 2\), then \(2p + 1 = 5\), which is also prime. So 2 is a Sophie Germain prime.
If \(p = 3\), then \(2p + 1 = 7\), which is also prime. So 3 is an SG prime.
If \(p = 5\), then \(2p + 1 = 11\), which is also prime. So 5 is an SG prime.
If \(p = 7\), then \(2p + 1 = 15\), which is NOT prime. So 7 is NOT an SG prime.
We are looking for the product of all the possible units digits of SG primes greater than 5.
Let's first narrow down the possible units digits of ALL primes greater than 5. First of all, no prime greater than 2 can end in an even digit (0, 2, 4, 6, 8), since no even number greater than 2 is prime. Likewise, no prime greater than 5 can end in 5, since all integers with a units digit of 5 are divisible by 5. So we can see that primes greater than 5 must end in 1, 3, 7, or 9 (e.g., 11, 13, 17, and 19). The question is whether the SG condition further restricts the possible units digits.
If a prime \(p\) ends in 1, then \(2p + 1\) ends in 3, which is a safe units digit for a prime.
If a prime \(p\) ends in 3, then \(2p + 1\) ends in 7, which is also a safe units digit for a prime.
If a prime \(p\) ends in 7, then \(2p + 1\) ends in 5, which is NOT a safe units digit for a prime greater than 5. So, in fact, no SG prime greater than 5 can end in 7.
Finally, if a prime \(p\) ends in 9, then \(2p + 1\) also ends in 9, which is a safe units digit for a prime.
Thus, SG primes can only end in 1, 3, or 9.
To be rigorous, we should find at least one example of an actual SG prime with each of those units digits, to confirm that we haven't missed something.
If \(p = 11\), then \(2p + 1 = 23\), which is also prime. So 11 is an SG prime, and we have 1 confirmed as a possible SG units digit.
If \(p = 13\), then \(2p + 1 = 27\), which is NOT prime. So 13 is NOT an SG prime.
Let's try 23:
If \(p = 23\), then \(2p + 1 = 47\), which is also prime. So 23 is an SG prime, and we have 3 confirmed as a possible SG units digit.
Finally, let's check 9 as a units digit.
If \(p = 19\), then \(2p + 1 = 39\), which is NOT prime. So 19 is NOT an SG prime.
But if \(p = 29\), then \(2p + 1 = 59\), which IS prime. So 29 is an SG prime, and we have 9 confirmed as a possible SG units digit.
The product of 1, 3, and 9 is 27.
Answer: D
30 Dec 2014, 08:13
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Hey,
I started calculating by 5, as an example to understand what is actually happening, and then went the other way around.
2(5)+1=10+1=11
So, I tested 189, going backwards: 189=188+1=2(94)+1. However, this ends up in 94, which is not a prime.
I went to the second highest, 27= 26+1= (13)2 +1. Since 13 is a prime, I accepted answer D.
Was this approach correct?
I started calculating by 5, as an example to understand what is actually happening, and then went the other way around.
2(5)+1=10+1=11
So, I tested 189, going backwards: 189=188+1=2(94)+1. However, this ends up in 94, which is not a prime.
I went to the second highest, 27= 26+1= (13)2 +1. Since 13 is a prime, I accepted answer D.
Was this approach correct?
