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Consecutive Prime numbers greater than equal to given number.
• Difficulty Level : Medium
• Last Updated : 05 May, 2021

### Question:

Given a number n, the task is to find two consecutive prime such that the product of these two prime is greater than or equal to n.

### Example:

Input: 14
Output: 3 5
Explanation: 3 and 5 are consecutive prime numbers whose product is greater than 14.

### Approach:

Suppose n is of the range 10^8 to 10^10. We cannot find out primes using sieve because the range is upto10^10.

We can find the required consecutive primes by doing the following method.

• Find the greatest prime which is less than sqrt(n) and store it in a temporary variable (first).
• Find the smallest prime which is greater than sqrt(n) and store it in a temporary variable(second).
• If the product of first and second is greater than equal to n then print it.
• Else find a prime just greater than second and print it along with second.

## C++

 `//C++ program for the above approach``#include ``#define endl "\n"``#define ll long long``using` `namespace` `std;` `//Function to check prime.``bool` `is_prime(ll n)``{``    ``if` `(n == 1)``    ``{``        ``return` `false``;``    ``}``    ``for` `(ll i = 2; i <= ``sqrt``(n); i++)``    ``{``        ``if` `(n % i == 0)``        ``{``          ``// It means it is not``          ``// a prime``            ``return` `false``;``        ``}``    ``}``  ``// No factor other than 1``  ``// therefore prime number``    ``return` `true``;``}` `//Function to find out the required``//consecutive primes.``void` `consecutive_primes(``int` `n)``{``    ``ll first = -1, second = -1;``  ` `    ``//Finding first prime just``    ``// less than sqrt(n).``    ``for` `(ll i = ``sqrt``(n); i >= 2; i--)``    ``{``        ``if` `(is_prime(i))``        ``{``            ``first = i;``            ``break``;``        ``}``    ``}``    ``// Finding prime just greater``    ``//than sqrt(n).``    ``for` `(ll i = ``sqrt``(n) + 1;``         ``i <= n / 2; i++)``    ``{``        ``if` `(is_prime(i))``        ``{``            ``second = i;``            ``break``;``        ``}``    ``}``    ``// Product of both prime is greater``    ``// than n then print it``    ``if` `(first * second >= n)``    ``{``        ``cout << first << ``" "``          ``<
Output
`3 5`

Time Complexity: O(sqrt(n))

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