The stronger Goldbach conjecture states that any even integer greater

Question

The stronger Goldbach conjecture states that any even integer greater than 7 can be written as the sum of two different prime numbers. For such representations of the even number 126, the largest possible difference between the two primes is

(A) 112
(B) 100
(C) 92
(D) 88
(E) 80

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paragw · Answer

Let the two prime numbers be a & b
We have 
a+b = 126 
and a-b = n, where n is the maximum difference between two primes which will add and give 126
Now subtract both the equations, we'll get
2b = 126-n
b = (126-n)/2
Now check the options one by one in decreasing order to get the answer

1. n=112
b = (126-112)/2 = 7
and a = 126-7 = 119, which is  not  a prime,  Rejected

2. n=100
 b = (126-100)/2 = 13
and a = 126-13 = 113, which is a  prime  and the largest difference between a & b
Therefore,  option B  is the correct answer

Dereno · Answer

Even integer greater than 7 can be represented as sum of two prime numbers.

8 = 3 + 5

Let the two prime numbers be p and a respectivel.

p + q = 126

p- q = options

2p = 126 + options

Substitute the options to find p and q,

A) 2p = 126 + 112

P = 119 , is divisible by 7. Hence, NOT a prime.

B) 2p = 126+100

p = 113 , which is a prime number.

113 - q = 100

q = 13.

Both,  p =113 and q =13  are Prime numbers.

Hence,  Option B

Diegoram17 · Answer

very clear explanation, thanks bro

Souradeep17 · Answer

We can try to find out the largest prime number which is less than 126, we see that the number is 113 . Here we have the sum of the prime numbers as 126 hence the other number should be 13. The difference between them is 100 which is option B

CareerLauncherCL · Answer

Here is the solution of the question, so let's understand it in a better way:-

We want two different primes p < q with p + q = 126 and the difference q - p as large as possible. To maximize the difference , make the smaller prime number p as small as possible while 126 - p is prime.

Now we are going to test small primes (to compute each complement):-
 
  p = 2 → 126 − 2 = 124 not prime. 
 p = 3 → 126 − 3 = 123 not prime. 
 p = 5 → 126 − 5 = 121=11×11 not prime. 
 p = 7 → 126−7 = 119 = 7×17 not prime. 
 p = 11 → 126−11=115=5×23 not prime. 
 p = 13 → 126−13 = 113. Check 113: it has no divisors up to \sqrt{}113(which is about 10.6), and it is a known prime.

So the first valid pair is 13 and 113. Their difference is 113−13 = 100.

So, the final correct answer would be :- Option (B) 100

MDB001 · Answer

Is this thread still active? I would like to ask a question. Bunuel

Bunuel · Answer

Yes, every topic on the forum to which you can reply is active. You can post your question in any thread you’re interested in.

MDB0001 · Answer

E is a 700-digit even number. Do we know any way to find p and q for this number? Of course p and q are primes and p + q = E.

MDB0001 · Answer

Bunuel  ? Any thoughts .

The stronger Goldbach conjecture states that any even integer greater

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The stronger Goldbach conjecture states that any even integer greater

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