# Smallest divisor D of N such that gcd(D, M) is greater than 1

Given two positive integers N and M., The task is to find the smallest divisor D of N such that gcd(D, M) > 1. If there are no such divisors, then print -1.

Examples:

Input: N = 8, M = 10
Output: 2

Input: N = 8, M = 1
Output: -1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A naive approach is to iterate for every factor and calculate the gcd of the factor and M. If it exceeds M, then we have the answer.

Time Complexity: O(N * log max(N, M))

An efficient approach is to iterate till sqrt(n) and check for gcd(i, m). If gcd(i, m) > 1, then we print and break it, else we check for gcd(n/i, m) and store the minimal of them.

Below is the implementation of the above approach.

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the minimum divisor ` `int` `findMinimum(``int` `n, ``int` `m) ` `{ ` `    ``int` `mini = m; ` ` `  `    ``// Iterate for all factors of N ` `    ``for` `(``int` `i = 1; i * i <= n; i++) { ` `        ``if` `(n % i == 0) { ` `            ``int` `sec = n / i; ` ` `  `            ``// Check for gcd > 1 ` `            ``if` `(__gcd(m, i) > 1) { ` `                ``return` `i; ` `            ``} ` ` `  `            ``// Check for gcd > 1 ` `            ``else` `if` `(__gcd(sec, m) > 1) { ` `                ``mini = min(sec, mini); ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// If gcd is m itself ` `    ``if` `(mini == m) ` `        ``return` `-1; ` `    ``else` `        ``return` `mini; ` `} ` `// Drivers code ` `int` `main() ` `{ ` `    ``int` `n = 8, m = 10; ` `    ``cout << findMinimum(n, m); ` `    ``return` `0; ` `} `

 `// Java implementation of the above approach ` `class` `GFG ` `{ ` ` `  `static` `int` `__gcd(``int` `a, ``int` `b)  ` `{  ` `    ``if` `(b == ``0``)  ` `        ``return` `a;  ` `    ``return` `__gcd(b, a % b);  ` `     `  `}  ` ` `  `// Function to find the minimum divisor ` `static` `int` `findMinimum(``int` `n, ``int` `m) ` `{ ` `    ``int` `mini = m; ` ` `  `    ``// Iterate for all factors of N ` `    ``for` `(``int` `i = ``1``; i * i <= n; i++) ` `    ``{ ` `        ``if` `(n % i == ``0``) ` `        ``{ ` `            ``int` `sec = n / i; ` ` `  `            ``// Check for gcd > 1 ` `            ``if` `(__gcd(m, i) > ``1``)  ` `            ``{ ` `                ``return` `i; ` `            ``} ` ` `  `            ``// Check for gcd > 1 ` `            ``else` `if` `(__gcd(sec, m) > ``1``) ` `            ``{ ` `                ``mini = Math.min(sec, mini); ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// If gcd is m itself ` `    ``if` `(mini == m) ` `        ``return` `-``1``; ` `    ``else` `        ``return` `mini; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``int` `n = ``8``, m = ``10``; ` `    ``System.out.println(findMinimum(n, m)); ` `} ` `} ` ` `  `// This code is contributed by chandan_jnu `

 `# Python3 implementation of the above approach ` `import` `math ` ` `  `# Function to find the minimum divisor  ` `def` `findMinimum(n, m):  ` ` `  `    ``mini, i ``=` `m, ``1` `     `  `    ``# Iterate for all factors of N  ` `    ``while` `i ``*` `i <``=` `n:  ` `        ``if` `n ``%` `i ``=``=` `0``:  ` `            ``sec ``=` `n ``/``/` `i  ` ` `  `            ``# Check for gcd > 1  ` `            ``if` `math.gcd(m, i) > ``1``:  ` `                ``return` `i  ` ` `  `            ``# Check for gcd > 1  ` `            ``elif` `math.gcd(sec, m) > ``1``:  ` `                ``mini ``=` `min``(sec, mini)  ` `             `  `        ``i ``+``=` `1` ` `  `    ``# If gcd is m itself  ` `    ``if` `mini ``=``=` `m: ` `        ``return` `-``1` `    ``else``: ` `        ``return` `mini  ` ` `  `# Drivers code  ` `if` `__name__ ``=``=` `"__main__"``:  ` ` `  `    ``n, m ``=` `8``, ``10` `    ``print``(findMinimum(n, m))  ` ` `  `# This code is contributed by Rituraj Jain `

 `// C# implementation of the above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `static` `int` `__gcd(``int` `a, ``int` `b)  ` `{  ` `    ``if` `(b == 0)  ` `        ``return` `a;  ` `    ``return` `__gcd(b, a % b);  ` `     `  `}  ` ` `  `// Function to find the minimum divisor ` `static` `int` `findMinimum(``int` `n, ``int` `m) ` `{ ` `    ``int` `mini = m; ` ` `  `    ``// Iterate for all factors of N ` `    ``for` `(``int` `i = 1; i * i <= n; i++) ` `    ``{ ` `        ``if` `(n % i == 0) ` `        ``{ ` `            ``int` `sec = n / i; ` ` `  `            ``// Check for gcd > 1 ` `            ``if` `(__gcd(m, i) > 1)  ` `            ``{ ` `                ``return` `i; ` `            ``} ` ` `  `            ``// Check for gcd > 1 ` `            ``else` `if` `(__gcd(sec, m) > 1) ` `            ``{ ` `                ``mini = Math.Min(sec, mini); ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// If gcd is m itself ` `    ``if` `(mini == m) ` `        ``return` `-1; ` `    ``else` `        ``return` `mini; ` `} ` ` `  `// Driver code ` `static` `void` `Main() ` `{ ` `    ``int` `n = 8, m = 10; ` `    ``Console.WriteLine(findMinimum(n, m)); ` `} ` `} ` ` `  `// This code is contributed by chandan_jnu `

1
if (__gcd(\$m, \$i) > 1)
{
return \$i;
}

// Check for gcd > 1
else if (__gcd(\$sec, \$m) > 1)
{
\$mini = min(\$sec, \$mini);
}
}
}

// If gcd is m itself
if (\$mini == \$m)
return -1;
else
return \$mini;
}

// Driver code
\$n = 8; \$m = 10;
echo(findMinimum(\$n, \$m));

// This code is contributed by Code_Mech.

Output:
```2
```

Time Complexity: O(sqrt N * log max(N, M))

Striver(underscore)79 at Codechef and codeforces D

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