Question

If x is a natural number and 3x-4, 4x-5, 5x-3 are three prime number then maximum value of 6x+5​

Answer 1

Given :- If x is a natural number and 3x-4, 4x-5, 5x-3 are three prime number then maximum value of 6x+5 ?

Solution :-

putting value of x one by one we get,

at x = 1 ,

• 3x - 4 => 3 * 1 - 4 = 3 - 4 = (-1) ≠ prime number .

at x = 2 ,

• 3x - 4 => 3 * 2 - 4 = 6 - 4 = 2 = prime number .
• 4x - 5 => 4 * 2 - 5 = 8 - 5 = 3 = prime number .
• 5x - 3 => 5 * 2 - 3 = 10 - 3 = 7 = prime number .
• all satisfy .

checking next also, since we need maximum value,

at x = 3 ,

• 3x - 4 => 3 * 3 - 4 = 9 - 4 = 5 = prime number .
• 4x - 5 => 4 * 3 - 5 = 12 - 5 = 7 = prime number .
• 5x - 3 => 5 * 3 - 3 = 15 - 3 = 12 ≠ prime number .

at x = 4 ,

• 3x - 4 => 3 * 4 - 4 = 12 - 4 = 8 ≠ prime number .

therefore, we can conclude that, value of x is 2 .

hence,

→ 6x + 5 = 6 * 2 + 5 = 12 + 5 = 17 (Ans.)

[Note :- Reason why values greater than 4 also not possible .

A prime number greater except 2 or 3 can be written in the form of 6q - 1 and 6q + 1 .

taking 6q - 1,

→ 3x - 4 = 6q - 1

→ 3x = 6q + 3

→ x = (2q + 1)

now when we put x in 5x - 3,

→ 5(2q + 1) - 3 = 10q + 5 - 3 = 10q + 2 = 2(5q + 1) => it becomes a even number . since 2 is only even prime number . This case is not possible .

taking 6q + 1 ,

→ 3x - 4 = 6q + 1 ,

→ 3x = 6q + 5

→ x = (6q + 5)/3

→ x = (6q/3) + (5/3)

for all values of q , RHS will not be a natural number as given that x is a natural number . Then, this is also not possible

Hence, we can conclude that, x = 2 satisfy only satisfy this condition .

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Answer 2

Given : x is a natural number

3x-4, 4x-5, 5x-3 are three prime number  

To Find :  maximum value of 6x+5​

Solution:

3x-4, 4x-5, 5x-3 are three prime number  

Without losing generality  assume that

x = 2q , 2q + 1  where q is integer

case 1 : x = 2q

3x - 4 = 3(2q)- 4    = 6q -  4 =  2(3q - 2)   Even number

only even prime number is  2

Hence 2(3q - 2)  = 2

=> 3q - 2 = 1

=> 3q = 3

=> q = 1

for q = 1  , x  = 2

3x - 4  = 2 ,  

4x - 5 = 3

5x - 3  = 7

all are even numbers

Case 2 :  x = 2q+1

3x-4  = 6q  - 1

4x-5  = 8q - 1

5x-3  = 10q + 2  = 2(5q + 1)   Even number

and only prime even number is 2

Hence  2(5q + 1)  = 2

=> 5q + 1 = 1

=> q = 0

Hence x = 1

3x - 4  = - 1  and 4x-5 = - 1   not   natural numbers

Hence not possible

so only possible combination is

x = 2

and prime numbers   2, 3 , 7

only possible value of 6x + 5  = 6(2) + 5  = 17

Hence maximum value of 6x + 5  = 17

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