S96-07

Question

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

Bunuel · Answer

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

pacifist85 · Answer

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?

viktorija · Answer

You should re-read the problem. You misunderstood what the question is actually asking.
The problem is asking you to find all the possible Sophie Germain prime units digits and calculate their product.

If you take a look at the solution, it shows very well what kind of numbers can be called Sophie Germain primes and what their possible units digits are.

Bunuel · Answer

Check alternative solution below. Hope it helps. 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 prime number greater than 5 can have only the following four units digits: 1, 3, 7, or 9. If the units digit of p is 1 then the units digit of 2p+1 would be 3, which is a possible units digit for a prime. For example consider p=11=prime --> 2p+1=23=prime; If the units digit of p is 3 then the units digit of 2p+1 would be 7, which is a possible units digit for a prime. For example consider p=23=prime --> 2p+1=47=prime; If the units digit of p is 7 then the units digit of 2p+1 would be 5, which is NOT a possible units digit for a prime; If the units digit of p is 9 then the units digit of 2p+1 would be 9, which is a possible units digit for a prime. For example consider p=29=prime --> 2p+1=59=prime. The product of all the possible units digits of Sophie Germain primes greater than 5 is 1*3*9=27. Answer: D. Check similar questions here: special-numbers-and-sequences-problems-174994.html

Umang Agarwal · Answer

p=53;2p+1=107 also satisfies the above criteria.
hence answer=81
3^2*9*1 = 81

Bunuel · Answer

Please re-read the question and the solutions. We need the product of all the possible units digits of Sophie Germain primes. The possible units digits are 1, 3 and 9. You should not use any of those more than once.

altairahmad · Answer

I think I didn't get the question.

Bunuel

Aren't there infinite (or a lot of) Sophie Germain primes ? e.g >5 are 11, 23, 29, 41, 53, 83, 89, 113, 131. The units digits of these few primes is 6561.

While there is a restriction on the smallest prime, there is no restriction on the greatest prime.

Why are we limiting our calculations to the units digits of first 3 primes only ?

Please help.

Bunuel · Answer

The product of all the possible units digits of Sophie Germain primes greater than 5 is Since the possible units digits of Sophie Germain primes greater than 5 are 1, 3, and 9, then the product of all the possible units digits of Sophie Germain primes greater than 5 is 1*3*9=27. I tried to explain this question here: https://gmatclub.com/forum/s96-184678.html#p1465189 Hope it helps.

akadiyan · Answer

I started manually validating primes and SP for primes greater than 5 and ended up 2 prime numbers 23 and 29 whose value for 2p+1 is also a prime number.

p = 23, sg prime = 47 --> unit digits of p = 3
p = 29 , sg prime = 59 --> unit digits of p = 9
p = 41 , sg prime = 83 --> unit digits of p = 1

For prime numbers>41 that satisifies the sg prime condition , the unit digits is always 1.

So the product of unit digits is 3*9*1 = 27

Ans:  D

SS97 · Answer

Hi  Bunuel  thank you for the explanation, this was really useful! I had one quick question challenging the statement that if the units digit of the Sophie prime numbers are 1,3 and 9 the product's unit digit might be ending with 7.

Taking an example:
The Sophie prime numbers are 11, 23, 29, 41, 43... now just these numbers we know the units digit of their product is 1 (1*3*9 -> 7*1*3). So how do we know if the units digit of the product will always be 7, it might depend on the number of digits right (in this case there were 5 digits whereas in some other case it might be 7 digits or so on). What am I missing here?

bumpbot · Answer

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