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# Count of numbers upto M divisible by given Prime Numbers

• Last Updated : 19 Mar, 2021

Given an array arr[] of Prime Numbers and a number M, the task is to count the number of elements in the range [1, M] that are divisible by any of the given prime numbers.

Examples:

Input: arr[] = {2, 3, 5, 7} M = 100
Output: 78
Explanation:
In total there are 78 numbers that are divisible by either of 2 3 5 or 7.

Input: arr[] = {2, 5, 7, 11} M = 200
Output: 137
Explanation:
In total there are 137 numbers that are divisible by either of 2 5 7 or 11.

Naive Approach: The idea is iterate over the range [1, M] and check if any of the array element is divides the element in the range [1, M] then count the element else check for the next number in the range.
Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to count the numbers that``// are divisible by the numbers in``// the array from range 1 to M``int` `count(``int` `a[], ``int` `M, ``int` `N)``{``    ``// Initialize the count variable``    ``int` `cnt = 0;` `    ``// Iterate over [1, M]``    ``for` `(``int` `i = 1; i <= M; i++) {` `        ``// Iterate over array elements arr[]``        ``for` `(``int` `j = 0; j < N; j++) {` `            ``// Check if i is divisible by a[j]``            ``if` `(i % a[j] == 0) {` `                ``// Increment the count``                ``cnt++;``                ``break``;``            ``}``        ``}``    ``}` `    ``// Return the answer``    ``return` `cnt;``}` `// Driver code``int` `main()``{``    ``// Given array arr[]``    ``int` `arr[] = { 2, 3, 5, 7 };` `    ``// Given Number M``    ``int` `m = 100;``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``// Function Call``    ``cout << count(arr, m, n);``    ``return` `0;``}`

## Java

 `// Java program for the above approach``class` `GFG{` `// Function to count the numbers that``// are divisible by the numbers in``// the array from range 1 to M``static` `int` `count(``int` `a[], ``int` `M, ``int` `N)``{``    ` `    ``// Initialize the count variable``    ``int` `cnt = ``0``;` `    ``// Iterate over [1, M]``    ``for``(``int` `i = ``1``; i <= M; i++)``    ``{``        ` `        ``// Iterate over array elements arr[]``        ``for``(``int` `j = ``0``; j < N; j++)``        ``{``            ` `            ``// Check if i is divisible by a[j]``            ``if` `(i % a[j] == ``0``)``            ``{``                ` `                ``// Increment the count``                ``cnt++;``                ``break``;``            ``}``        ``}``    ``}``    ` `    ``// Return the answer``    ``return` `cnt;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ` `    ``// Given array arr[]``    ``int` `arr[] = { ``2``, ``3``, ``5``, ``7` `};` `    ``// Given number M``    ``int` `m = ``100``;``    ``int` `n = arr.length;` `    ``// Function call``    ``System.out.print(count(arr, m, n));``}``}` `// This code is contributed by Amit Katiyar`

## Python3

 `# Python3 program for the above approach` `# Function to count the numbers that``# are divisible by the numbers in``# the array from range 1 to M``def` `count(a, M, N):``    ` `    ``# Initialize the count variable``    ``cnt ``=` `0` `    ``# Iterate over [1, M]``    ``for` `i ``in` `range``(``1``, M ``+` `1``):` `        ``# Iterate over array elements arr[]``        ``for` `j ``in` `range``(N):` `            ``# Check if i is divisible by a[j]``            ``if` `(i ``%` `a[j] ``=``=` `0``):` `                ``# Increment the count``                ``cnt ``+``=` `1``                ``break` `    ``# Return the answer``    ``return` `cnt` `# Driver code` `# Given list lst``lst ``=` `[ ``2``, ``3``, ``5``, ``7` `]` `# Given number M``m ``=` `100``n ``=` `len``(lst)` `# Function call``print``(count(lst, m, n))` `# This code is contributed by vishu2908   `

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{` `// Function to count the numbers that``// are divisible by the numbers in``// the array from range 1 to M``static` `int` `count(``int` `[]a, ``int` `M, ``int` `N)``{``    ` `    ``// Initialize the count variable``    ``int` `cnt = 0;` `    ``// Iterate over [1, M]``    ``for``(``int` `i = 1; i <= M; i++)``    ``{``        ` `        ``// Iterate over array elements []arr``        ``for``(``int` `j = 0; j < N; j++)``        ``{``            ` `            ``// Check if i is divisible by a[j]``            ``if` `(i % a[j] == 0)``            ``{``                ` `                ``// Increment the count``                ``cnt++;``                ``break``;``            ``}``        ``}``    ``}``    ` `    ``// Return the answer``    ``return` `cnt;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ` `    ``// Given array []arr``    ``int` `[]arr = { 2, 3, 5, 7 };` `    ``// Given number M``    ``int` `m = 100;``    ``int` `n = arr.Length;` `    ``// Function call``    ``Console.Write(count(arr, m, n));``}``}` `// This code is contributed by Amit Katiyar`

## Javascript

 ``
Output
`78`

Time Complexity: O(N*M)
Auxiliary Space: O(1)
Another Approach: Another method to solve this problem is use Dynamic Programming and Seive. Mark all the numbers up to M that are divisible by any prime number in the array. Then count all the marked numbers and print it.

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