Given an integer **N**, the task is to count the number of distinct prime triplets **(a, b, c)** from the range **[1, N]** such that **a < b < c ≤ N** and **a + b = c**.**Note:** Two prime tuples are distinct if at least one of the primes present in them are different.

**Examples:**

Input:N = 6Output:1Explanation:Among numbers in the range [1, 6], the only prime triplet is (2, 3, 5) (Since 2 + 3 = 5).

Input:N = 10Output:2Explanation:The distinct prime triplets satisfying the condition are (2, 3, 5), (2, 5, 7).

**Approach:** The problem can be solved based on the observation stated below:

**Observation:**

For every prime number

pfrom1toN, it is a part of a triplet if and only if it can be represented as a sum of two prime numbers.

Since a prime number is an odd number, it must be equal to the sum of an even number and an odd number.

Hence the only even prime is2. Therefore, for a prime numberpto constitute a unique tuple (2, p-2, p), the numberp – 2must be a prime number.

Follow the steps below to solve the problem:

- Initialize a variable, say
**count = 0**, to store the number of prime triplets. - Iterate from
**1**to**N**and for each number**p**, check if this number**p**and**p – 2**are prime or not. - If they are prime, then increment
**count**by**1**. - Print the value of
**count**.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to check if a` `// number is a prime or not` `bool` `isPrime(` `int` `N)` `{` ` ` `if` `(N <= 1)` ` ` `return` `false` `;` ` ` `for` `(` `int` `i = 2; i <= ` `sqrt` `(N); i++) {` ` ` `if` `(N % i == 0)` ` ` `return` `false` `;` ` ` `}` ` ` `return` `true` `;` `}` `// Function to count the number` `// of valid prime triplets` `void` `countPrimeTuples(` `int` `N)` `{` ` ` `// Stores the count` ` ` `// of prime triplets` ` ` `int` `count = 0;` ` ` `// Iterate from 2 to N and check for each` ` ` `// p, whether p & (p - 2) are prime or not` ` ` `for` `(` `int` `i = 2; i <= N; i++) {` ` ` `if` `(isPrime(i) && isPrime(i - 2))` ` ` `count++;` ` ` `}` ` ` `// Print the count obtained` ` ` `cout << count;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 6;` ` ` `countPrimeTuples(N);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.io.*;` `class` `GFG` `{` ` ` ` ` `// Function to check if a` ` ` `// number is a prime or not` ` ` `static` `boolean` `isPrime(` `int` `N)` ` ` `{` ` ` `if` `(N <= ` `1` `)` ` ` `return` `false` `;` ` ` `for` `(` `int` `i = ` `2` `; i <= Math.sqrt(N); i++)` ` ` `{` ` ` `if` `(N % i == ` `0` `)` ` ` `return` `false` `;` ` ` `}` ` ` `return` `true` `;` ` ` `}` ` ` `// Function to count the number` ` ` `// of valid prime triplets` ` ` `static` `void` `countPrimeTuples(` `int` `N)` ` ` `{` ` ` ` ` `// Stores the count` ` ` `// of prime triplets` ` ` `int` `count = ` `0` `;` ` ` `// Iterate from 2 to N and check for each` ` ` `// p, whether p & (p - 2) are prime or not` ` ` `for` `(` `int` `i = ` `2` `; i <= N; i++)` ` ` `{` ` ` `if` `(isPrime(i) && isPrime(i - ` `2` `))` ` ` `count++;` ` ` `}` ` ` `// Print the count obtained` ` ` `System.out.println(count);` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main (String[] args)` ` ` `{ ` ` ` `int` `N = ` `6` `;` ` ` `countPrimeTuples(N);` ` ` `}` `}` `// This code is contributed by Dharanendra L V.` |

## Python3

`# Python3 program for the above approach` `import` `math` `# Function to check if a` `# number is a prime or not` `def` `isPrime(N) :` ` ` `if` `(N <` `=` `1` `) :` ` ` `return` `False` ` ` `for` `i ` `in` `range` `(` `2` `, ` `int` `(math.sqrt(N) ` `+` `1` `)):` ` ` `if` `(N ` `%` `i ` `=` `=` `0` `) :` ` ` `return` `False` ` ` `return` `True` `# Function to count the number` `# of valid prime triplets` `def` `countPrimeTuples(N) :` ` ` ` ` `# Stores the count` ` ` `# of prime triplets` ` ` `count ` `=` `0` ` ` `# Iterate from 2 to N and check for each` ` ` `# p, whether p & (p - 2) are prime or not` ` ` `for` `i ` `in` `range` `(` `2` `, N ` `+` `1` `):` ` ` `if` `(isPrime(i) ` `and` `isPrime(i ` `-` `2` `)) :` ` ` `count ` `+` `=` `1` ` ` ` ` `# Prthe count obtained` ` ` `print` `(count)` `# Driver Code` `N ` `=` `6` `countPrimeTuples(N)` `# This code is contributed by susmitakundugoaldanga.` |

## C#

`// C# program for the above approach` `using` `System;` `public` `class` `GFG` `{` ` ` ` ` `// Function to check if a` ` ` `// number is a prime or not` ` ` `static` `bool` `isPrime(` `int` `N)` ` ` `{` ` ` `if` `(N <= 1)` ` ` `return` `false` `;` ` ` `for` `(` `int` `i = 2; i <= Math.Sqrt(N); i++)` ` ` `{` ` ` `if` `(N % i == 0)` ` ` `return` `false` `;` ` ` `}` ` ` `return` `true` `;` ` ` `}` ` ` `// Function to count the number` ` ` `// of valid prime triplets` ` ` `static` `void` `countPrimeTuples(` `int` `N)` ` ` `{` ` ` ` ` `// Stores the count` ` ` `// of prime triplets` ` ` `int` `count = 0;` ` ` `// Iterate from 2 to N and check for each` ` ` `// p, whether p & (p - 2) are prime or not` ` ` `for` `(` `int` `i = 2; i <= N; i++)` ` ` `{` ` ` `if` `(isPrime(i) && isPrime(i - 2))` ` ` `count++;` ` ` `}` ` ` `// Print the count obtained` ` ` `Console.WriteLine(count);` ` ` `}` ` ` `// Driver Code` ` ` `static` `public` `void` `Main ()` ` ` `{` ` ` `int` `N = 6;` ` ` `countPrimeTuples(N);` ` ` `}` `}` `// This code is contributed by Dharanendra L V.` |

**Output:**

1

**Time Complexity: **O(N^{3/2})**Auxiliary Space: **O(1)

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