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# S96-07

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:50
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Difficulty:

95% (hard)

Question Stats:

32% (01:35) correct 68% (01:55) wrong based on 152 sessions

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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

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16 Sep 2014, 00:50
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.

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Status: Math is psycho-logical
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30 Dec 2014, 07:13
2
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.

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04 Jan 2015, 14:53
pacifist85 wrote:
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.

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.
Math Expert
Joined: 02 Sep 2009
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05 Jan 2015, 04:58
1
pacifist85 wrote:
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.

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.

Check similar questions here: special-numbers-and-sequences-problems-174994.html
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11 Dec 2016, 13:50
Is it a question which can be solved in 2 minutes? I took almost 5 minutes to substitute values and validate. I'd get really worried of i see a question like this on the screen on the day of the exam.
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Joined: 02 Sep 2009
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11 Dec 2016, 14:25
vasundhara88 wrote:
Is it a question which can be solved in 2 minutes? I took almost 5 minutes to substitute values and validate. I'd get really worried of i see a question like this on the screen on the day of the exam.

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.

Hope it's clear.
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25 Dec 2016, 00:22
p=53;2p+1=107 also satisfies the above criteria.
3^2*9*1 = 81
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25 Dec 2016, 01:12
Umang Agarwal wrote:
p=53;2p+1=107 also satisfies the above criteria.
3^2*9*1 = 81

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.
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14 May 2017, 01:40
I think that it is not clear enough that we need to use each digit exactly once...

It is worth to mention..
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29 Aug 2017, 12:04
I think this is a poor-quality question and I don't agree with the explanation. Some examples of SG primes are 23, 47, 59 etc.
Whereas in explanation and calculation, unit digit of SG prime are considered as 1, 3 and 9 which is wrong. Unit digit of SG prime can only be 3, 7 and 9 and therefore, product of 3,7 and 9= 189 must be the answer.
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29 Aug 2017, 21:07
kritika24agarwal wrote:
I think this is a poor-quality question and I don't agree with the explanation. Some examples of SG primes are 23, 47, 59 etc.
Whereas in explanation and calculation, unit digit of SG prime are considered as 1, 3 and 9 which is wrong. Unit digit of SG prime can only be 3, 7 and 9 and therefore, product of 3,7 and 9= 189 must be the answer.

You did not read the question and solution carefully. 47 is NO a “Sophie Germain” becasue 2*47 + 1 = 95, which is NOT a prime. We are told that "a “Sophie Germain” prime is any positive prime number p for which 2p + 1 is also prime.".

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05 Dec 2017, 13:31
primes greater than 5 Makes all the difference in the world
Re: S96-07 &nbs [#permalink] 05 Dec 2017, 13:31
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# S96-07

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