# Find numbers which are multiples of first array and factors of second array

Given two arrays A[] and B[], the task is to find the integers which are divisible by all the elements of array A[] and divide all the elements of array B[].

Examples:

Input: A[] = {1, 2, 2, 4}, B[] = {16, 32, 64}
Output: 4 8 16
4, 8 and 16 are the only numbers that
are multiples of all the elements of array A[]
and divide all the elements of array B[]

Input: A[] = {2, 3, 6}, B[] = {42, 84}
Output: 6 42

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

Approach: If X is a multiple of all the elements of the first array then X must be a multiple of the LCM of all the elements of the first array.
Similarly, If X is a factor of all the elements of the second array then it must be a factor of the GCD of all the elements of the second array and such X will exist only if GCD of the second array is divisible by the LCM of the first array.
If it is divisible then X can be any value from the range [LCM, GCD] which is a multiple of LCM and evenly divides GCD.

Below is the implementation of above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the LCM of two numbers ` `int` `lcm(``int` `x, ``int` `y) ` `{ ` `    ``int` `temp = (x * y) / __gcd(x, y); ` `    ``return` `temp; ` `} ` ` `  `// Function to print the requried numbers ` `void` `findNumbers(``int` `a[], ``int` `n, ``int` `b[], ``int` `m) ` `{ ` ` `  `    ``// To store the lcm of array a[] elements ` `    ``// and the gcd of array b[] elements ` `    ``int` `lcmA = 1, gcdB = 0; ` ` `  `    ``// Finding LCM of first array ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``lcmA = lcm(lcmA, a[i]); ` ` `  `    ``// Finding GCD of second array ` `    ``for` `(``int` `i = 0; i < m; i++) ` `        ``gcdB = __gcd(gcdB, b[i]); ` ` `  `    ``// No such element exists ` `    ``if` `(gcdB % lcmA != 0) { ` `        ``cout << ``"-1"``; ` `        ``return``; ` `    ``} ` ` `  `    ``// All the multiples of lcmA which are ` `    ``// less than or equal to gcdB and evenly ` `    ``// divide gcdB will satisfy the conditions ` `    ``int` `num = lcmA; ` `    ``while` `(num <= gcdB) { ` `        ``if` `(gcdB % num == 0) ` `            ``cout << num << ``" "``; ` `        ``num += lcmA; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `a[] = { 1, 2, 2, 4 }; ` `    ``int` `b[] = { 16, 32, 64 }; ` ` `  `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` `    ``int` `m = ``sizeof``(b) / ``sizeof``(b); ` ` `  `    ``findNumbers(a, n, b, m); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `static` `int` `__gcd(``int` `a, ``int` `b)  ` `{  ` `    ``if` `(b == ``0``)  ` `        ``return` `a;  ` `    ``return` `__gcd(b, a % b);  ` `     `  `} ` ` `  `// Function to return the LCM of two numbers ` `static` `int` `lcm(``int` `x, ``int` `y) ` `{ ` `    ``int` `temp = (x * y) / __gcd(x, y); ` `    ``return` `temp; ` `} ` ` `  `// Function to print the requried numbers ` `static` `void` `findNumbers(``int` `a[], ``int` `n,  ` `                        ``int` `b[], ``int` `m) ` `{ ` ` `  `    ``// To store the lcm of array a[] elements ` `    ``// and the gcd of array b[] elements ` `    ``int` `lcmA = ``1``, gcdB = ``0``; ` ` `  `    ``// Finding LCM of first array ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``lcmA = lcm(lcmA, a[i]); ` ` `  `    ``// Finding GCD of second array ` `    ``for` `(``int` `i = ``0``; i < m; i++) ` `        ``gcdB = __gcd(gcdB, b[i]); ` ` `  `    ``// No such element exists ` `    ``if` `(gcdB % lcmA != ``0``)  ` `    ``{ ` `        ``System.out.print(``"-1"``); ` `        ``return``; ` `    ``} ` ` `  `    ``// All the multiples of lcmA which are ` `    ``// less than or equal to gcdB and evenly ` `    ``// divide gcdB will satisfy the conditions ` `    ``int` `num = lcmA; ` `    ``while` `(num <= gcdB)  ` `    ``{ ` `        ``if` `(gcdB % num == ``0``) ` `            ``System.out.print(num + ``" "``); ` `        ``num += lcmA; ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `a[] = { ``1``, ``2``, ``2``, ``4` `}; ` `    ``int` `b[] = { ``16``, ``32``, ``64` `}; ` ` `  `    ``int` `n = a.length; ` `    ``int` `m = b.length; ` ` `  `    ``findNumbers(a, n, b, m); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 implementation of the approach  ` `from` `math ``import` `gcd ` ` `  `# Function to return the LCM of two numbers  ` `def` `lcm( x, y) : ` `     `  `    ``temp ``=` `(x ``*` `y) ``/``/` `gcd(x, y);  ` `    ``return` `temp;  ` ` `  `# Function to print the requried numbers  ` `def` `findNumbers(a, n, b, m) :  ` ` `  `    ``# To store the lcm of array a[] elements  ` `    ``# and the gcd of array b[] elements  ` `    ``lcmA ``=` `1``; __gcdB ``=` `0``;  ` ` `  `    ``# Finding LCM of first array  ` `    ``for` `i ``in` `range``(n) :  ` `        ``lcmA ``=` `lcm(lcmA, a[i]);  ` ` `  `    ``# Finding GCD of second array  ` `    ``for` `i ``in` `range``(m) :  ` `        ``__gcdB ``=` `gcd(__gcdB, b[i]);  ` ` `  `    ``# No such element exists  ` `    ``if` `(__gcdB ``%` `lcmA !``=` `0``) : ` `        ``print``(``"-1"``);  ` `        ``return``;  ` ` `  `    ``# All the multiples of lcmA which are  ` `    ``# less than or equal to gcdB and evenly  ` `    ``# divide gcdB will satisfy the conditions  ` `    ``num ``=` `lcmA;  ` `    ``while` `(num <``=` `__gcdB) : ` `        ``if` `(__gcdB ``%` `num ``=``=` `0``) : ` `            ``print``(num, end ``=` `" "``);  ` `             `  `        ``num ``+``=` `lcmA;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``a ``=` `[ ``1``, ``2``, ``2``, ``4` `]; ` `    ``b ``=` `[ ``16``, ``32``, ``64` `]; ` `     `  `    ``n ``=` `len``(a); ` `    ``m ``=` `len``(b); ` `     `  `    ``findNumbers(a, n, b, m);  ` `     `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `static` `int` `__gcd(``int` `a, ``int` `b)  ` `{  ` `    ``if` `(b == 0)  ` `        ``return` `a;  ` `    ``return` `__gcd(b, a % b);  ` `} ` ` `  `// Function to return the LCM of two numbers ` `static` `int` `lcm(``int` `x, ``int` `y) ` `{ ` `    ``int` `temp = (x * y) / __gcd(x, y); ` `    ``return` `temp; ` `} ` ` `  `// Function to print the requried numbers ` `static` `void` `findNumbers(``int` `[]a, ``int` `n,  ` `                        ``int` `[]b, ``int` `m) ` `{ ` ` `  `    ``// To store the lcm of array a[] elements ` `    ``// and the gcd of array b[] elements ` `    ``int` `lcmA = 1, gcdB = 0; ` ` `  `    ``// Finding LCM of first array ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``lcmA = lcm(lcmA, a[i]); ` ` `  `    ``// Finding GCD of second array ` `    ``for` `(``int` `i = 0; i < m; i++) ` `        ``gcdB = __gcd(gcdB, b[i]); ` ` `  `    ``// No such element exists ` `    ``if` `(gcdB % lcmA != 0)  ` `    ``{ ` `        ``Console.Write(``"-1"``); ` `        ``return``; ` `    ``} ` ` `  `    ``// All the multiples of lcmA which are ` `    ``// less than or equal to gcdB and evenly ` `    ``// divide gcdB will satisfy the conditions ` `    ``int` `num = lcmA; ` `    ``while` `(num <= gcdB)  ` `    ``{ ` `        ``if` `(gcdB % num == 0) ` `            ``Console.Write(num + ``" "``); ` `        ``num += lcmA; ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]a = { 1, 2, 2, 4 }; ` `    ``int` `[]b = { 16, 32, 64 }; ` ` `  `    ``int` `n = a.Length; ` `    ``int` `m = b.Length; ` ` `  `    ``findNumbers(a, n, b, m); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```4 8 16
```

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