A “Sophie Germain” prime is any positive prime number p for

Why 7 is not considered for the final answer?

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 greater than 5.

It might be simple, but I have a doubt here. The question asks product of all the possible unit digits of Sophie Germain primes.

As 47 is a sophie germain prime number and prime number and 47 is > than 5, so 7 being the unit digit should be included in the product to get the final answer. That is why I marked 189.

A “Sophie Germain” prime is any positive prime number p for which 2p + 1 is also prime. 47 is NOT a “Sophie Germain” prime because 2p+1=95, which is NOT a prime. Again, a “Sophie Germain” prime cannot have 7 as its units digit because the units digit of 2p+1 in this case would be 5. No prime greater than 5 has 5 as its units digit.

Hope it's clear.

Thanks Banuel.

I read the premise wrongly.

I see - I misunderstood the question at first...

It's asking for all of the possible UNIQUE units digits, not all of the units digits of SG primes multiplied out.

For example, 11, 23, 29, 53... are primes. Unique units are only 1, 3, and 9, which multiply to 27.

If interpreted literally, it would go on forever (i.e. answer = infinite).

Knowing this and the answer choices, there's only one intended Q&A combo possible - unique units and not infinite.

I do think, however, that they should have worded it more clearly as it is ambiguous.

VeritasKarishma Bunuel chetan2u egmat AjiteshArun

taking prime number as p=23 , its 2p+1 =47 and 47 is a prime number.

Then why are we not considering unit digit of 47 (i.e 7) here.

The question says:

A “Sophie Germain” prime is any positive prime number p for which 2p + 1 is also prime.

This means that it is 23 that is a Sophie Germain prime, not 47, because;

if p=23, 2p+1 is prime
but
if p=47, 2p+1 is not prime

Therefore, we won't consider 47 a Sophie Germain prime.

Only p is the Sophie Germain prime, not the corresponding 2p + 1.

It's strange to ask for the "product of all the possible units digits" here, because then there's no reason to bother checking numbers with a units digit of "1" -- a product will be the same whether we include "1" or not. So you can save yourself a quarter of the work here and only check 3, 7 and 9, which was done perfectly in other solutions above.

Here's my take.

Since the question talks about "Sophie Germain" Primes we have to check for same.
2, 3, 5, 11, 23, 29, 41, 53, 71, 83, 89 and so on are all "Sophie Germain" Primes. For these primes greater than 5 units digits are 1,3,9 which repeats, hence product is 1*3*9=27.

Answer (D)

Rule: the only Possible Units Digits that a Prime Number > 5 can have are the following: 1 - 3 - 7 - or 9

2p + 1 ALSO must = a Prime No.

Just using the Units Digits of the Possible Prime Numbers after applying 2p + 1:

Prime No. with a Units Digit of 1: the Units Digit = 2(......1) + 1 = ......2 + 2 = ......3

There must exist a Prime No. in the Infinite Amount of Primes that ends in a 1 and that is also a "Sophie Gemaine" Prime

(in fact 11 works: 2 * 11 + 1 = 23, another Prime Number)


Prime No. with a Units Digit of 3: the Units Digit = 2* (.....3) + 1 = ....6 + 1 = 7 Units Digit
There will be a Sophie-Germain Prime with a Units Digit of 3.


Prime No. with a Units Digit of 7: the Units Digit = 2 * (.....7) + 1 = .....4 + 1 = 5 Units Digit
The Number will be Divisible by 5, because the Units Digit will end in a 5. Thus, Any Primes > 5 that end in a Units Digit of 7 will never be a Sophie-Germain Prime

Prime No. with a Units Digit of 9: the Units Digit = 2 * (....9) + 1 = .....8 + 1 = 9 Units Diigt
There will be a Sophie-Germain Prime with a Units Digit of 9


The Product of all the Possible Units Digits = 1 * 3 * 9 = 27

Answer -D-

how do we know that there is no other sophie prime after 29? I mean there are infinitive prime numbers?

Yes, there is an infinite number of prime numbers. And there are lots of Sophie Germain primes larger than 29 -- for example, 41 is another.

No one knows for sure if there are infinitely many Sophie Germain primes; that's a famous unsolved problem in mathematics. Mathematicians suspect there are infinitely many, because they aren't so rare that one would expect there to be a finite number of them, but no one has proved that yet.

Fortunately in this question, you don't need to establish how many Sophie Germain primes there are. The question only asks about their possible units digits, so you only need to consider the very finite set of possible units digits a number can have.

actually the questions asks abou "The product of all the possible units digits of Sophie Germain primes greater than 5 is" and we can't know this for sure. The question should ask about "A possible product of the unit digits.."

I think you might be interpreting the question to mean something more like "what is the product of the units digits of all Sophie Germain primes greater than 5", and if the question asked that, it would be unanswerable -- no one even knows if there's a finite number of that type of prime. But the question asks something different; it asks for the product of the possible units digits of those primes. So we just need to consider each possible units digit from 0 through 9, work out if each is a possible units digit of a Sophie Germain prime, and multiply together those that we find are possible.

I do find it strange to ask for a product in this question though, as I mentioned in an earlier post.

But when we take 23 , we get 47. Both are prime...so condition is satisfied...then how u can use the logic 7*2+1 = 5. ? I'm not getting it

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