# Count numbers less than N whose modulo with A is equal to B

Given three non-negative integers **A**, **B**, and **N** where **A** is *non-zero*, the task is to find the number of integers less than or equal to **N** whose modulo with **A** gives the value **B**.

**Examples:**

Input:A = 6, B = 3, N = 15Output:3Explanation:The numbers 3, 9 and 15 are less than or equal to N (= 15) and their modulo with A (= 6) is equal to B (= 3). Therefore, the count such numbers is 3.

Input:A = 4, B = 1, C = 8Output:2

**Approach:** The given problem can be solved by using an observation based on mathematics. Follow the steps below to solve the problem:

- If the value of
**B**is**at least A**, then print**0**, as there can’t be any such numbers whose modulo with**A**gives the value**B**. - Otherwise, If the value of
**B**is**0**, then print the value of**C / A**as the count of such numbers whose modulo with**A**gives the value**B**. - Otherwise, perform the following steps:
- Initialize a variable, say ans, with the floor value of
**C / A**. - If the value of
**(ans * A + B)**is less than or equal to**N**, then increment the value of**ans**by**1**. - After completing the above steps, print the value of
**ans**as the count of such numbers whose modulo with**A**gives the value**B**.

- Initialize a variable, say ans, with the floor value of

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to count numbers less than` `// N, whose modulo with A gives B` `void` `countValues(` `int` `A, ` `int` `B, ` `int` `C)` `{` ` ` `// If the value of B at least A` ` ` `if` `(B >= A) {` ` ` `cout << 0;` ` ` `return` `;` ` ` `}` ` ` `// If the value of B is 0 or not` ` ` `if` `(B == 0) {` ` ` `cout << C / A;` ` ` `return` `;` ` ` `}` ` ` `// Stores the resultant count` ` ` `// of numbers less than N` ` ` `int` `ans = C / A;` ` ` `if` `(ans * A + B <= C) {` ` ` `// Update the value of ans` ` ` `ans++;` ` ` `}` ` ` `// Print the value of ans` ` ` `cout << ans;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `A = 6, B = 3, N = 15;` ` ` `countValues(A, B, N);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `public` `class` `MyClass` `{` ` ` `// Function to count numbers less than` `// N, whose modulo with A gives B` `static` `void` `countValues(` `int` `A, ` `int` `B, ` `int` `C)` `{` ` ` ` ` `// If the value of B at least A` ` ` `if` `(B >= A) {` ` ` `System.out.println(` `0` `);` ` ` `return` `;` ` ` `}` ` ` `// If the value of B is 0 or not` ` ` `if` `(B == ` `0` `) {` ` ` `System.out.println(C / A);` ` ` `return` `;` ` ` `}` ` ` `// Stores the resultant count` ` ` `// of numbers less than N` ` ` `int` `ans = C / A;` ` ` `if` `(ans * A + B <= C) {` ` ` `// Update the value of ans` ` ` `ans++;` ` ` `}` ` ` `// Print the value of ans` ` ` `System.out.println(ans);` `}` `// Driver Code` `public` `static` `void` `main(String args[])` `{` ` ` `int` `A = ` `6` `, B = ` `3` `, N = ` `15` `;` ` ` `countValues(A, B, N);` `} ` `}` `// This code in contributed by SoumikMondal` |

## Python3

`# Python3 program for the above approach` `# Function to count numbers less than` `# N, whose modulo with A gives B` `def` `countValues(A, B, C):` ` ` ` ` `# If the value of B at least A` ` ` `if` `(B >` `=` `A):` ` ` `print` `(` `0` `)` ` ` `return` ` ` `# If the value of B is 0 or not` ` ` `if` `(B ` `=` `=` `0` `):` ` ` `print` `(C ` `/` `/` `A)` ` ` `return` ` ` `# Stores the resultant count` ` ` `# of numbers less than N` ` ` `ans ` `=` `C` `/` `/` `A` ` ` `if` `(ans ` `*` `A ` `+` `B <` `=` `C):` ` ` `# Update the value of ans` ` ` `ans ` `+` `=` `1` ` ` `# Print the value of ans` ` ` `print` `(ans)` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `A ` `=` `6` ` ` `B ` `=` `3` ` ` `N ` `=` `15` ` ` ` ` `countValues(A, B, N)` `# This code is contributed by SURENDRA_GANGWAR` |

## C#

`// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{` ` ` `// Function to count numbers less than` `// N, whose modulo with A gives B` `static` `void` `countValues(` `int` `A, ` `int` `B, ` `int` `C)` `{` ` ` ` ` `// If the value of B at least A` ` ` `if` `(B >= A)` ` ` `{` ` ` `Console.Write(0);` ` ` `return` `;` ` ` `}` ` ` `// If the value of B is 0 or not` ` ` `if` `(B == 0)` ` ` `{` ` ` `Console.Write(C / A);` ` ` `return` `;` ` ` `}` ` ` `// Stores the resultant count` ` ` `// of numbers less than N` ` ` `int` `ans = C / A;` ` ` `if` `(ans * A + B <= C)` ` ` `{` ` ` ` ` `// Update the value of ans` ` ` `ans++;` ` ` `}` ` ` `// Print the value of ans` ` ` `Console.Write(ans);` `}` `// Driver code` `public` `static` `void` `Main()` `{` ` ` `int` `A = 6, B = 3, N = 15;` ` ` `countValues(A, B, N);` `}` `}` `// This code is contributed by sanjoy_62` |

## Javascript

`<script>` `// JavaScript program for the above approach` `// Function to count numbers less than` `// N, whose modulo with A gives B` `function` `countValues(A, B, C)` `{` ` ` `// If the value of B at least A` ` ` `if` `(B >= A) {` ` ` `document.write(0);` ` ` `return` `;` ` ` `}` ` ` `// If the value of B is 0 or not` ` ` `if` `(B == 0) {` ` ` `document.write(parseInt(C / A));` ` ` `return` `;` ` ` `}` ` ` `// Stores the resultant count` ` ` `// of numbers less than N` ` ` `let ans = parseInt(C / A);` ` ` `if` `(ans * A + B <= C) {` ` ` `// Update the value of ans` ` ` `ans++;` ` ` `}` ` ` `// Print the value of ans` ` ` `document.write(ans);` `}` `// Driver Code` ` ` `let A = 6, B = 3, N = 15;` ` ` `countValues(A, B, N);` `</script>` |

**Output:**

3

**Time Complexity:** O(1)**Auxiliary Space:** O(1)

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