Sum of Some Primes

Note that $x>y$ , since $x-y$ is positive. Since $x$ and $y$ are both prime, this means that $x$ must be greater than 2 and therefore odd. If $y$ were odd, $x+y$ would be an even number greater than 2 and hence not prime. Thus, $y$ must be even, i.e., $y=2$ . Now we want an odd prime $x$ such that $x-2$ and $x+2$ are both prime. In other words, we want three consecutive odd numbers that are all prime. But one of $x-2$ , $x$ , and $x+2$ is divisible by 3, so in order to be a prime it must be 3 . Clearly that one must be $x-2$ , the smallest of the three numbers, and we have our unique solution: $x=5$ and $y=2$ , and $x+y+(x+y)+(x-y)=3x+y=17$ .

[LaTeX edits - Calvin]

Since x, y, x-y and x+y are all positive prime integers, then these four primes are greater than or equal to 2. Thus x>y, since x-y is a positive prime.

One of them must be an even prime integer, since even + odd = odd. So let y= 2, since 2 is the only even prime integer.

Hence, x+2 and x-2 are prime too! This means that x-2, x, and x+2 are 3 consecutive primes with a difference of 2.

Hence, x=5 and the four primes are 2,3,5 and 7!

Thus, 2+3+5+7=17!

Ans. 17

:)

since x and y are both positive numbers ,

x−y is positive prime integer(given) ,

therefore, x-y > 0
i.e. x>y
and x and y - both prime number (given)

also x+y is a positive prime number

now consider

Case1-

if x and y are both odd prime numbers then their sum can never be a prime number (sum of two odd numbers is always an even number and except 2 no even prime number exists) [i.e. REJECTED]

Case2-

Out of x and y one number should be even,
(because all the prime numbers are odd except 2 and given the sum of x and y should be a prime i.e their sum=odd )

Now, we know 2 is the smallest and only even prime number , x > y and x+y=odd

hence y=2

now if we check we'll find three numbers in series that is having a common difference of two amongst them i.e. x-y , x , x+y = x-2 , x , x+2
(satisfying the condition of three numbers in series which are twin primes)

hence 3,5,7 is the only existing series of twin prime pattern therefore by logic

x-2=3

x=5

x+2=7

and x=5 satisfies all the three equations hence we got the solutions as x=5 and y=2 therefore x=5

y=2

x-y=3

x+y=7

their sum is
x + y+ x-y + x+y = 3x+y = 3*5 + 2 = 15+2 = 17

ans ----- 17

Note that x-y>0 so x>y. Since both of x and y is prime, it means that x must greater than 2 or therefore odd. If y were an odd number, x+y would be an even number which is not satisfy x+y is a prime. Hence, y must be 2.

Now we just have to find three consecutive odd numbers that are all prime and formed x-2, x, x+2. But in that sequence, one of them is divisible by 3, in order to be a prime. Look that there are some integers x, that x=3n, x=3n+1, or x=3n+2. x won't be 3n since x is prime, then if x=3n+1, x+2=3(n+1) and if x=3n+2, x-2=3n which both x+2 and x-2 in the first and second case, respectively, are not prime. So in every case, x-2, x, or x+2 can be a multiple of 3. Since 3-2 is not prime, hence, x-2=3 (other multiplication of 3 are not prime) that force x=5 and y=2. So x+y+(x-y)+(x+y)=3x+y=17

Noticed that,

For $x+y$ is a prime number, $x$ and $y$ must be an odd number and even number respestively. Among the prime numbers, only $2$ is an even number. As $x\neq2$ ( $2$ is the smallest prime number, it would be false as $x-y$ is a prime number), so $y=2$ . Other than that, for $x-y\geq3$ , and it is a prime number, the minimum value for $x$ is $5$ .

Now the sum of 4 values is

$(x)+(y)+(x-y)+(x+y) =3x+y =3(5)+(2) =17$

List down the first few prime factors, which are 2,3,5,7,11...etc

Since x-y, x must be bigger than y. x cannot be 3 since 3-2=1, which is not a prime number. When x=5, y can be 2 or 3. However, x+y must be a prime number, and 5+3=8, which is not a prime number. Therefore y=2.

The sum of all integers=5+2+(5-2)+(5+2)=17

x=5 which is prime y=2 which is prime which implies x-y=3 which is also prime and x+y=7 which is also prime. Hence x+y+x-y+x+y= 5+2+3+7=17

One of the numbers has to be 2, since its the only even prime.

Using trial and error, it becomes evident that x=5 and y=2.

Any other value for x will not give a prime when added to 2 or when 2 is subtracted from it.

Therefore x=5, y=2, x+y=7and x-y=3. The sum of which is 17.

usually x+y is an even number, the only possible that x+y is a prime integer is y=2, and thus x=5.

Since the problem requires positive prime integers, we start looking at the list of prime numbers starting with 2, 3, 5, and 7. Since x-y should be positive, y must be greater than x. Also, 1 should not be the result of x-y, so we can start with x=5, and y could be 2 to obtain another prime which is 3. Further, x + y would give another prime , 7.

x=5; y=2; x+y=7 ; x-y=3 All 4 integers are prime

x=5 y=2 X-Y=3 X+Y=7

5+2+3+7=17

first try any prime no. so that x,y,x+y,x-y is prime

for eg ,let x=5 and y=2; x-y=3and x+y=7;

 whole four no. is prime
 adding prime no. we get 
   5+2+3+7=17

ANS is 17

We are given that $x > 0$ , $y > 0$ and $x - y > 0 \Rightarrow x > y$ .

Suppose $x$ and $y$ are both odd then $x - y$ and $x + y$ are even. There is only one even prime thus $x$ and $y$ cannot both be odd. One of $x$ and $y$ must be even, since $2$ is the smallest prime (and even), we let $y = 2$ .

The other three integers are $x - 2$ , $x$ and $x + 2$ . These three numbers are three consecutive odd numbers, thus one of them must be divisible by $3$ . Since $x - 2$ , $x$ and $x+2$ are prime, we must have $x - 2 = 3$ as it is the smallest odd prime available. Thus $x = 5$ . Therefore we have that the four integers are $5, 2, 3$ and $7$ . Hence the sum is $2 + 3 + 5 + 7 = 17$ .

looking at x + y, since it should be prime, x and y should be an even and an odd numbers, thus it must be 2 for x or y

x=5,y=2 x-y=3 x+y=7 =5+2+3+7=17

Since the sum of any two prime number except 2 is even. Therefore x+y is not a prime number

Therefore x=2 (or) y=2

since given that x-y is a prime number

therefore y=2 and x=5 ->by trial and error method.

therefore sum of all the prime numbers is 5+2+5-2+5+2 =17

Y is greater than x so that x-y is positive (assuming that in general, prime numbers are not negative). Try some numbers and it turns out that x=3 and y=2 works perfectly to satisfy the condition.

So x cannot be 3 as x-y will be 1, which is not as per the question. So x will be 5. Check for all the other values it will satisfy the conditions.