Given four integers **A, B, C** and **D**. The task is to find the count of integers in the range **[A, B]** that are not divisible by **C** and **D** .

**Examples:**

Input:A = 4, B = 9, C = 2, D = 3Output:2

5 and 7 are such integers.

Input:A = 10, B = 50, C = 4, D = 6Output:28

**Approach:** First include all the integers in the range in the required answer i.e. **B – A + 1**. Then remove all the numbers which are divisible by **C** and **D** and finally add all the numbers which are divisible by both **C** and **D**.

Below is the implementation of the above approach:

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; `
` ` `// Function to return the count of ` `// integers from the range [a, b] that ` `// are not divisible by c and d ` `int` `countNums(` `int` `a, ` `int` `b, ` `int` `c, ` `int` `d) `
`{ ` ` ` `// Numbers which are divisible by c `
` ` `int` `x = b / c - (a - 1) / c; `
` ` ` ` `// Numbers which are divisible by d `
` ` `int` `y = b / d - (a - 1) / d; `
` ` ` ` `// Find lowest common factor of c and d `
` ` `int` `k = (c * d) / __gcd(c, d); `
` ` ` ` `// Numbers which are divisible by both c and d `
` ` `int` `z = b / k - (a - 1) / k; `
` ` ` ` `// Return the required answer `
` ` `return` `b - a + 1 - x - y + z; `
`} ` ` ` `// Driver code ` `int` `main() `
`{ ` ` ` `int` `a = 10, b = 50, c = 4, d = 6; `
` ` ` ` `cout << countNums(a, b, c, d); `
` ` ` ` `return` `0; `
`} ` |

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`// Java implementation of the approach ` `class` `GFG `
`{ ` ` ` `// Function to return the count of ` `// integers from the range [a, b] that ` `// are not divisible by c and d ` `static` `int` `countNums(` `int` `a, ` `int` `b, ` `int` `c, ` `int` `d) `
`{ ` ` ` `// Numbers which are divisible by c `
` ` `int` `x = b / c - (a - ` `1` `) / c; `
` ` ` ` `// Numbers which are divisible by d `
` ` `int` `y = b / d - (a - ` `1` `) / d; `
` ` ` ` `// Find lowest common factor of c and d `
` ` `int` `k = (c * d) / __gcd(c, d); `
` ` ` ` `// Numbers which are divisible by both c and d `
` ` `int` `z = b / k - (a - ` `1` `) / k; `
` ` ` ` `// Return the required answer `
` ` `return` `b - a + ` `1` `- x - y + z; `
`} ` `static` `int` `__gcd(` `int` `a, ` `int` `b) `
`{ ` ` ` `if` `(b == ` `0` `) `
` ` `return` `a; `
` ` `return` `__gcd(b, a % b); `
`} ` ` ` `// Driver code ` `public` `static` `void` `main(String []args) `
`{ ` ` ` `int` `a = ` `10` `, b = ` `50` `, c = ` `4` `, d = ` `6` `; `
` ` ` ` `System.out.println(countNums(a, b, c, d)); `
`} ` `} ` ` ` `// This code is contributed by PrinciRaj1992 ` |

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`# Python3 implementation of the approach ` `from` `math ` `import` `gcd `
` ` `# Function to return the count of ` `# integers from the range [a, b] that ` `# are not divisible by c and d ` `def` `countNums(a, b, c, d) : `
` ` ` ` `# Numbers which are divisible by c `
` ` `x ` `=` `b ` `/` `/` `c ` `-` `(a ` `-` `1` `) ` `/` `/` `c; `
` ` ` ` `# Numbers which are divisible by d `
` ` `y ` `=` `b ` `/` `/` `d ` `-` `(a ` `-` `1` `) ` `/` `/` `d; `
` ` ` ` `# Find lowest common factor of c and d `
` ` `k ` `=` `(c ` `*` `d) ` `/` `/` `gcd(c, d); `
` ` ` ` `# Numbers which are divisible `
` ` `# by both c and d `
` ` `z ` `=` `b ` `/` `/` `k ` `-` `(a ` `-` `1` `) ` `/` `/` `k; `
` ` ` ` `# Return the required answer `
` ` `return` `(b ` `-` `a ` `+` `1` `-` `x ` `-` `y ` `+` `z); `
` ` `# Driver code ` `if` `__name__ ` `=` `=` `"__main__"` `: `
` ` ` ` `a ` `=` `10` `; b ` `=` `50` `; c ` `=` `4` `; d ` `=` `6` `; `
` ` ` ` `print` `(countNums(a, b, c, d)); `
` ` `# This code is contributed by AnkitRai01 ` |

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`// C# implementation of the approach ` `using` `System; `
` ` `class` `GFG `
`{ ` ` ` `// Function to return the count of ` `// integers from the range [a, b] that ` `// are not divisible by c and d ` `static` `int` `countNums(` `int` `a, ` `int` `b, ` `int` `c, ` `int` `d) `
`{ ` ` ` `// Numbers which are divisible by c `
` ` `int` `x = b / c - (a - 1) / c; `
` ` ` ` `// Numbers which are divisible by d `
` ` `int` `y = b / d - (a - 1) / d; `
` ` ` ` `// Find lowest common factor of c and d `
` ` `int` `k = (c * d) / __gcd(c, d); `
` ` ` ` `// Numbers which are divisible by both c and d `
` ` `int` `z = b / k - (a - 1) / k; `
` ` ` ` `// Return the required answer `
` ` `return` `b - a + 1 - x - y + z; `
`} ` ` ` `static` `int` `__gcd(` `int` `a, ` `int` `b) `
`{ ` ` ` `if` `(b == 0) `
` ` `return` `a; `
` ` `return` `__gcd(b, a % b); `
`} ` ` ` `// Driver code ` `public` `static` `void` `Main(String []args) `
`{ ` ` ` `int` `a = 10, b = 50, c = 4, d = 6; `
` ` ` ` `Console.WriteLine(countNums(a, b, c, d)); `
`} ` `} ` ` ` `// This code is contributed by Rajput-Ji ` |

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**Output:**

28

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