Given a positive integer N > 1. Find the maximum count of prime numbers whose sum is equal to given N.
Input : N = 5
Output : 2
Explanation : 2 and 3 are two prime numbers whose sum is 5.
Input : N = 6
Explanation : 2, 2, 2 are three prime numbers whose sum is 6.
For maximum number of primes whose sum is equal to given n, prime numbers must be as small as possible. So, 2 is smallest possible prime number and is an even number. Next prime number greater than 2 is 3 which is odd. So, for any given n there are two conditions, either n will be odd or even.
- Case 1 : n is even, In this case n/2 will be the answer (n/2 number of 2 will result into sum of n).
- Case 1 : n is odd, In this case floor(n/2) will be the answer ((n-3)/2 number of 2 and one 3 will result into sum of n.
Below is the implementation of the above approach:
- Minimum number of single digit primes required whose sum is equal to N
- Length of largest sub-array having primes strictly greater than non-primes
- Count primes that can be expressed as sum of two consecutive primes and 1
- Find three integers less than or equal to N such that their LCM is maximum
- Find four factors of N with maximum product and sum equal to N
- Find four factors of N with maximum product and sum equal to N | Set-2
- Find four factors of N with maximum product and sum equal to N | Set 3
- Find maximum product of digits among numbers less than or equal to N
- Maximum count of equal numbers in an array after performing given operations
- Circular primes less than n
- Palindromic Primes
- Product of Primes of all Subsets
- Count Primes in Ranges
- Alternate Primes till N
- Product of all primes in the range from L to R
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