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# Split n into maximum composite numbers

• Difficulty Level : Easy
• Last Updated : 15 Dec, 2022

Given n, print the maximum number of composite numbers that sum up to n. First few composite numbers are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, ………
Examples:

```Input: 90
Output: 22
Explanation: If we add 21 4's, then we
get 84 and then add 6 to it, we get 90.

Input: 10
Output: 2
Explanation: 4 + 6 = 10```

Below are some important observations.

1. If the number is less than 4, it won’t have any combinations.
2. If the number is 5, 7, 11, it wont have any splitting.
3. Since smallest composite number is 4, it makes sense to use maximum number of 4s.
4. For numbers that don’t leave a composite remainder when divided by 4, we do following. If remainder is 1, we subtract 9 from it to get the number which is perfectly divisible by 4. If the remainder is 2, then subtract 6 from it to make n a number which is perfectly divisible by 4. If the remainder is 3, then subtract 15 from it to make n perfectly divisible by 4, and 15 can be made up by 9 + 6.

So the main observation is to make n such that is composes of maximum no of 4’s and the remaining can be made up by 6 and 9. We won’t need composite numbers more than that, as the composite numbers above 9 can be made up of 4, 6, and 9.
Below is the implementation of the above approach

## C++

 `// CPP program to split a number into maximum` `// number of composite numbers.` `#include ` `using` `namespace` `std;`   `// function to calculate the maximum number of` `// composite numbers adding upto n` `int` `count(``int` `n)` `{` `    ``// 4 is the smallest composite number` `    ``if` `(n < 4)` `        ``return` `-1;`   `    ``// stores the remainder when n is divided` `    ``// by 4` `    ``int` `rem = n % 4;`   `    ``// if remainder is 0, then it is perfectly ` `    ``// divisible by 4.` `    ``if` `(rem == 0)` `        ``return` `n / 4;`   `    ``// if the remainder is 1` `    ``if` `(rem == 1) {`   `        ``// If the number is less than 9, that` `        ``// is 5, then it cannot be expressed as ` `        ``// 4 is the only composite number less ` `        ``// than 5` `        ``if` `(n < 9)` `            ``return` `-1;`   `        ``// If the number is greater than 8, and ` `        ``// has a remainder of 1, then express n` `        ``// as n-9 a and it is perfectly divisible ` `        ``// by 4 and for 9, count 1.` `        ``return` `(n - 9) / 4 + 1;` `    ``}`   `    `  `    ``// When remainder is 2, just subtract 6 from n, ` `    ``// so that n is perfectly divisible by 4 and` `    ``// count 1 for 6 which is subtracted.` `    ``if` `(rem == 2)` `        ``return` `(n - 6) / 4 + 1;`     `    ``// if the number is 7, 11 which cannot be ` `    ``// expressed as sum of any composite numbers` `    ``if` `(rem == 3) {` `        ``if` `(n < 15)` `            ``return` `-1;`   `        ``// when the remainder is 3, then subtract` `        ``// 15 from it and n becomes perfectly ` `        ``// divisible by 4 and we add 2 for 9 and 6,` `        ``// which is getting subtracted to make n ` `        ``// perfectly divisible by 4.` `        ``return` `(n - 15) / 4 + 2;` `    ``}` `}`   `// driver program to test the above function` `int` `main()` `{` `    ``int` `n = 90;` `    ``cout << count(n) << endl;`   `    ``n = 143;` `    ``cout << count(n) << endl;` `    ``return` `0; ` `}`

## Java

 `// Java program to split a number into maximum` `// number of composite numbers.` `import` `java.io.*;`   `class` `GFG ` `{` `    ``// function to calculate the maximum number of` `    ``// composite numbers adding upto n` `    ``static` `int` `count(``int` `n)` `    ``{` `        ``// 4 is the smallest composite number` `        ``if` `(n < ``4``)` `            ``return` `-``1``;` `    `  `        ``// stores the remainder when n is divided` `        ``// by 4` `        ``int` `rem = n % ``4``;` `    `  `        ``// if remainder is 0, then it is perfectly ` `        ``// divisible by 4.` `        ``if` `(rem == ``0``)` `            ``return` `n / ``4``;` `    `  `        ``// if the remainder is 1` `        ``if` `(rem == ``1``) {` `    `  `            ``// If the number is less than 9, that` `            ``// is 5, then it cannot be expressed as ` `            ``// 4 is the only composite number less ` `            ``// than 5` `            ``if` `(n < ``9``)` `                ``return` `-``1``;` `    `  `            ``// If the number is greater than 8, and ` `            ``// has a remainder of 1, then express n` `            ``// as n-9 a and it is perfectly divisible ` `            ``// by 4 and for 9, count 1.` `            ``return` `(n - ``9``) / ``4` `+ ``1``;` `        ``}` `    `  `        `  `        ``// When remainder is 2, just subtract 6 from n, ` `        ``// so that n is perfectly divisible by 4 and` `        ``// count 1 for 6 which is subtracted.` `        ``if` `(rem == ``2``)` `            ``return` `(n - ``6``) / ``4` `+ ``1``;` `    `  `    `  `        ``// if the number is 7, 11 which cannot be ` `        ``// expressed as sum of any composite numbers` `        ``if` `(rem == ``3``) ` `        ``{` `            ``if` `(n < ``15``)` `                ``return` `-``1``;` `    `  `            ``// when the remainder is 3, then subtract` `            ``// 15 from it and n becomes perfectly ` `            ``// divisible by 4 and we add 2 for 9 and 6,` `            ``// which is getting subtracted to make n ` `            ``// perfectly divisible by 4.` `            ``return` `(n - ``15``) / ``4` `+ ``2``;` `        ``}` `        ``return` `0``;` `    ``}` `    `  `    ``// Driver program ` `    ``public` `static` `void` `main (String[] args) ` `    ``{` `        ``int` `n = ``90``;` `        ``System.out.println(count(n));`   `        ``n = ``143``;` `        ``System.out.println(count(n));` `    ``}` `} `   `// This code is contributed by vt_m.`

## Python3

 `# Python3 program to split a number into ` `# maximum number of composite numbers.`   `# Function to calculate the maximum number ` `# of composite numbers adding upto n` `def` `count(n):`   `    ``# 4 is the smallest composite number` `    ``if` `(n < ``4``):` `        ``return` `-``1`   `    ``# stores the remainder when n  ` `    ``# is divided n is divided by 4` `    ``rem ``=` `n ``%` `4`   `    ``# if remainder is 0, then it is  ` `    ``# perfectly divisible by 4.` `    ``if` `(rem ``=``=` `0``):` `        ``return` `n ``/``/` `4`   `    ``# if the remainder is 1` `    ``if` `(rem ``=``=` `1``):`   `        ``# If the number is less than 9, that` `        ``# is 5, then it cannot be expressed as ` `        ``# 4 is the only composite number less ` `        ``# than 5` `        ``if` `(n < ``9``):` `            ``return` `-``1`   `        ``# If the number is greater than 8, and ` `        ``# has a remainder of 1, then express n` `        ``# as n-9 a and it is perfectly divisible ` `        ``# by 4 and for 9, count 1.` `        ``return` `(n ``-` `9``) ``/``/` `4` `+` `1` `    `    `    `  `    ``# When remainder is 2, just subtract 6 from n, ` `    ``# so that n is perfectly divisible by 4 and` `    ``# count 1 for 6 which is subtracted.` `    ``if` `(rem ``=``=` `2``):` `        ``return` `(n ``-` `6``) ``/``/` `4` `+` `1`     `    ``# if the number is 7, 11 which cannot be ` `    ``# expressed as sum of any composite numbers` `    ``if` `(rem ``=``=` `3``): ` `        ``if` `(n < ``15``):` `            ``return` `-``1`   `        ``# when the remainder is 3, then subtract` `        ``# 15 from it and n becomes perfectly ` `        ``# divisible by 4 and we add 2 for 9 and 6,` `        ``# which is getting subtracted to make n ` `        ``# perfectly divisible by 4.` `        ``return` `(n ``-` `15``) ``/``/` `4` `+` `2`   `# Driver Code` `n ``=` `90` `print``(count(n))`   `n ``=` `143` `print``(count(n))`   `# This code is contributed by Anant Agarwal.`

## C#

 `// C# program to split a number into maximum` `// number of composite numbers.` `using` `System;`   `class` `GFG {` `    `  `    ``// function to calculate the maximum number` `    ``// of composite numbers adding upto n` `    ``static` `int` `count(``int` `n)` `    ``{` `        `  `        ``// 4 is the smallest composite number` `        ``if` `(n < 4)` `            ``return` `-1;` `     `  `        ``// stores the remainder when n is divided` `        ``// by 4` `        ``int` `rem = n % 4;` `     `  `        ``// if remainder is 0, then it is perfectly ` `        ``// divisible by 4.` `        ``if` `(rem == 0)` `            ``return` `n / 4;` `     `  `        ``// if the remainder is 1` `        ``if` `(rem == 1) {` `     `  `            ``// If the number is less than 9, that` `            ``// is 5, then it cannot be expressed as ` `            ``// 4 is the only composite number less ` `            ``// than 5` `            ``if` `(n < 9)` `                ``return` `-1;` `     `  `            ``// If the number is greater than 8, and ` `            ``// has a remainder of 1, then express n` `            ``// as n-9 a and it is perfectly divisible ` `            ``// by 4 and for 9, count 1.` `            ``return` `(n - 9) / 4 + 1;` `        ``}` `     `  `         `  `        ``// When remainder is 2, just subtract 6 from n, ` `        ``// so that n is perfectly divisible by 4 and` `        ``// count 1 for 6 which is subtracted.` `        ``if` `(rem == 2)` `            ``return` `(n - 6) / 4 + 1;` `     `  `     `  `        ``// if the number is 7, 11 which cannot be ` `        ``// expressed as sum of any composite numbers` `        ``if` `(rem == 3) ` `        ``{` `            ``if` `(n < 15)` `                ``return` `-1;` `     `  `            ``// when the remainder is 3, then subtract` `            ``// 15 from it and n becomes perfectly ` `            ``// divisible by 4 and we add 2 for 9 and 6,` `            ``// which is getting subtracted to make n ` `            ``// perfectly divisible by 4.` `            ``return` `(n - 15) / 4 + 2;` `        ``}` `        `  `        ``return` `0;` `    ``}` `     `  `    ``// Driver program ` `    ``public` `static` `void` `Main() ` `    ``{` `        ``int` `n = 90;` `        ``Console.WriteLine(count(n));` ` `  `        ``n = 143;` `        ``Console.WriteLine(count(n));` `    ``}` `} ` ` `  `// This code is contributed by Anant Agarwal.`

## PHP

 ``

## Javascript

 ``

Output:

```22
34 ```

Time complexity: O(1)
Auxiliary Space: O(1)