# Divide N into K unique parts such that gcd of those parts is maximum

Given a positive integer **N**, the task is to divide it into **K** unique parts such that the sum of these parts is equal to the original number and the **gcd** of all the parts is maximum. Print the **maximum gcd** if such a division exists else print **-1**.

**Examples:**

Input:N = 6, K = 3

Output:1

Only possible division with unique

elements is (1, 2, 3) and gcd(1, 2, 3) = 1.

Input:N = 18, K = 3

Output:3

**Naive approach:** Find all the possible divisions of **N** and compute the maximum gcd of them. But this approach will take exponential time and space.

**Efficient approach:** Let the divisions of **N** be A_{1}, A_{2}……..A_{K}

Now, it is known that **gcd(a, b) = gcd(a, b, a + b)** and hence, **gcd(A _{1}, A_{2}….A_{K}) = gcd(A_{1}, A_{2}….A_{K}, A_{1} + A_{2}…. + A_{K})**

But

**A**and hence, the gcd of the divisions will be one of the factors of

_{1}+ A_{2}…. + A_{K}= N**N**.

Let

**a**be the factor of

**N**which can be the possible answer:

Since

**a**is the gcd, all the division will be the multiples of

**a**and hence, for

**K**unique

**B**,

_{i}**a * B**

_{1}+ a * B_{2}……. + a * B_{K}= N**a * (B**

_{1}+ B_{2}…….+ B_{K}) = NSince all the

**B**are unique,

_{i}**B**

_{1}+ B_{2}…….+ B_{K}≥ 1 + 2 + 3 ….. + K**B**

_{1}+ B_{2}…….+ B_{K}≥ K * (K + 1) / 2Hence, all the factors of

**N**whose complementary factor is greater than or equal to

**K * (K + 1) / 2**can be one of the possible answers and we have take to the maximum of all the possible answers.

Below is the implementation of the above approach:

## CPP

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to calculate maximum GCD ` `int` `maxGCD(` `int` `N, ` `int` `K) ` `{ ` ` ` ` ` `// Minimum possible sum for ` ` ` `// K unique positive integers ` ` ` `int` `minSum = (K * (K + 1)) / 2; ` ` ` ` ` `// It is not possible to divide ` ` ` `// N into K unique parts ` ` ` `if` `(N < minSum) ` ` ` `return` `-1; ` ` ` ` ` `// All the factors greater than sqrt(N) ` ` ` `// are complementary of the factors less ` ` ` `// than sqrt(N) ` ` ` `int` `i = ` `sqrt` `(N); ` ` ` `int` `res = 1; ` ` ` `while` `(i >= 1) { ` ` ` ` ` `// If i is a factor of N ` ` ` `if` `(N % i == 0) { ` ` ` `if` `(i >= minSum) ` ` ` `res = max(res, N / i); ` ` ` ` ` `if` `(N / i >= minSum) ` ` ` `res = max(res, i); ` ` ` `} ` ` ` `i--; ` ` ` `} ` ` ` ` ` `return` `res; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `N = 18, K = 3; ` ` ` ` ` `cout << maxGCD(N, K); ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the approach ` `import` `java.io.*; ` `import` `java.lang.Math; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to calculate maximum GCD ` ` ` `static` `int` `maxGCD(` `int` `N, ` `int` `K) ` ` ` `{ ` ` ` ` ` `// Minimum possible sum for ` ` ` `// K unique positive integers ` ` ` `int` `minSum = (K * (K + ` `1` `)) / ` `2` `; ` ` ` ` ` `// It is not possible to divide ` ` ` `// N into K unique parts ` ` ` `if` `(N < minSum) ` ` ` `return` `-` `1` `; ` ` ` ` ` `// All the factors greater than sqrt(N) ` ` ` `// are complementary of the factors less ` ` ` `// than sqrt(N) ` ` ` `int` `i = (` `int` `) Math.sqrt(N); ` ` ` `int` `res = ` `1` `; ` ` ` `while` `(i >= ` `1` `) ` ` ` `{ ` ` ` ` ` `// If i is a factor of N ` ` ` `if` `(N % i == ` `0` `) ` ` ` `{ ` ` ` `if` `(i >= minSum) ` ` ` `res = Math.max(res, N / i); ` ` ` ` ` `if` `(N / i >= minSum) ` ` ` `res = Math.max(res, i); ` ` ` `} ` ` ` `i--; ` ` ` `} ` ` ` ` ` `return` `res; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main (String[] args) ` ` ` `{ ` ` ` `int` `N = ` `18` `, K = ` `3` `; ` ` ` ` ` `System.out.println(maxGCD(N, K)); ` ` ` `} ` `} ` ` ` `// This code is contributed by ApurvaRaj ` |

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## Python

`# Python3 implementation of the approach ` `from` `math ` `import` `sqrt,ceil,floor ` ` ` `# Function to calculate maximum GCD ` `def` `maxGCD(N, K): ` ` ` ` ` `# Minimum possible sum for ` ` ` `# K unique positive integers ` ` ` `minSum ` `=` `(K ` `*` `(K ` `+` `1` `)) ` `/` `2` ` ` ` ` `# It is not possible to divide ` ` ` `# N into K unique parts ` ` ` `if` `(N < minSum): ` ` ` `return` `-` `1` ` ` ` ` `# All the factors greater than sqrt(N) ` ` ` `# are complementary of the factors less ` ` ` `# than sqrt(N) ` ` ` `i ` `=` `ceil(sqrt(N)) ` ` ` `res ` `=` `1` ` ` `while` `(i >` `=` `1` `): ` ` ` ` ` `# If i is a factor of N ` ` ` `if` `(N ` `%` `i ` `=` `=` `0` `): ` ` ` `if` `(i >` `=` `minSum): ` ` ` `res ` `=` `max` `(res, N ` `/` `i) ` ` ` ` ` `if` `(N ` `/` `i >` `=` `minSum): ` ` ` `res ` `=` `max` `(res, i) ` ` ` ` ` `i` `-` `=` `1` ` ` ` ` `return` `res ` ` ` `# Driver code ` ` ` `N ` `=` `18` `K ` `=` `3` ` ` `print` `(maxGCD(N, K)) ` ` ` `# This code is contributed by mohit kumar 29 ` |

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## C#

`// C# implementation of the approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to calculate maximum GCD ` ` ` `static` `int` `maxGCD(` `int` `N, ` `int` `K) ` ` ` `{ ` ` ` ` ` `// Minimum possible sum for ` ` ` `// K unique positive integers ` ` ` `int` `minSum = (K * (K + 1)) / 2; ` ` ` ` ` `// It is not possible to divide ` ` ` `// N into K unique parts ` ` ` `if` `(N < minSum) ` ` ` `return` `-1; ` ` ` ` ` `// All the factors greater than sqrt(N) ` ` ` `// are complementary of the factors less ` ` ` `// than sqrt(N) ` ` ` `int` `i = (` `int` `) Math.Sqrt(N); ` ` ` `int` `res = 1; ` ` ` `while` `(i >= 1) ` ` ` `{ ` ` ` ` ` `// If i is a factor of N ` ` ` `if` `(N % i == 0) ` ` ` `{ ` ` ` `if` `(i >= minSum) ` ` ` `res = Math.Max(res, N / i); ` ` ` ` ` `if` `(N / i >= minSum) ` ` ` `res = Math.Max(res, i); ` ` ` `} ` ` ` `i--; ` ` ` `} ` ` ` `return` `res; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `N = 18, K = 3; ` ` ` ` ` `Console.WriteLine(maxGCD(N, K)); ` ` ` `} ` `} ` ` ` `// This code is contributed by AnkitRai01 ` |

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**Output:**

3

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