Maximum positive integer divisible by C and is in the range [A, B]

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Given three positive integers A, B, and C. The task is to find the maximum integer X > 0 such that:

X % C = 0 and
X must belong to the range [A, B]

Print -1 if no such number i.e. X exists.

Examples:

Input: A = 2, B = 4, C = 2
Output: 4
B is itself divisible by C.

Input: A = 5, B = 10, C = 4
Output: 8 
B is not divisible by C. 
So maximum multiple of 4(C) smaller than 10(B) is 8

Approach:

If B is a multiple of C then B is the required number.
Else get the maximum multiple of C just lesser than B which is the required answer.

Below is the implementation of the above approach:

C++

// C++ implementation of the above approach
#include <iostream>
using namespace std;
 
// Function to return the required number
int getMaxNum(int a, int b, int c)
{
 
    // If b % c = 0 then b is the
    // required number
    if (b % c == 0)
        return b;
 
    // Else get the maximum multiple of
    // c smaller than b
    int x = ((b / c) * c);
     
    if (x >= a && x <= b)
        return x;
    else
        return -1;
}
 
// Driver code
int main()
{
    int a = 2, b = 10, c = 3;
    cout << getMaxNum(a, b, c);
    return 0;
}

Java

// Java implementation of the above approach
import java.io.*;
 
class GFG 
{
     
// Function to return the required number
static int getMaxNum(int a, int b, int c)
{
 
    // If b % c = 0 then b is the
    // required number
    if (b % c == 0)
        return b;
 
    // Else get the maximum multiple of
    // c smaller than b
    int x = ((b / c) * c);
     
    if (x >= a && x <= b)
        return x;
    else
        return -1;
}
 
// Driver code
public static void main (String[] args) 
{
    int a = 2, b = 10, c = 3;
    System.out.println(getMaxNum(a, b, c));
}
}
 
// This Code is contributed by ajit.. 

Python3

# Python3 implementation of the above approach
 
# Function to return the required number
def getMaxNum(a, b, c):
 
    # If b % c = 0 then b is the
    # required number
    if (b % c == 0):
        return b
 
    # Else get the maximum multiple 
    # of c smaller than b
    x = ((b //c) * c)
     
    if (x >= a and x <= b):
        return x
    else:
        return -1
 
# Driver code
a, b, c = 2, 10, 3
print(getMaxNum(a, b, c))
 
# This code is contributed 
# by Mohit Kumar

C#

// C# implementation of the above approach
using System;
class GFG 
{
     
// Function to return the required number
static int getMaxNum(int a, int b, int c)
{
 
    // If b % c = 0 then b is the
    // required number
    if (b % c == 0)
        return b;
 
    // Else get the maximum multiple of
    // c smaller than b
    int x = ((b / c) * c);
     
    if (x >= a && x <= b)
        return x;
    else
        return -1;
}
 
// Driver code
public static void Main () 
{
    int a = 2, b = 10, c = 3;
    Console.WriteLine(getMaxNum(a, b, c));
}
}
 
// This Code is contributed by Code_Mech.. 

PHP

<?php
// PHP implementation of the above approach
// Function to return the required number
function getMaxNum($a, $b, $c)
{
 
    // If b % c = 0 then b is the
    // required number
    if ($b % $c == 0)
        return $b;
 
    // Else get the maximum multiple
    // of c smaller than b
    $x = ((int)($b / $c) * $c);
     
    if ($x >= $a && $x <= $b)
        return $x;
    else
        return -1;
}
 
// Driver code
$a = 2; $b = 10; $c = 3;
echo(getMaxNum($a, $b, $c));
 
// This Code is contributed
// by Mukul Singh
?>

Javascript

<script>
 
// Javascript implementation of the above approach
 
// Function to return the required number
function getMaxNum(a, b, c)
{
 
    // If b % c = 0 then b is the
    // required number
    if (b % c == 0)
        return b;
 
    // Else get the maximum multiple of
    // c smaller than b
    var x = (parseInt(b / c) * c);
     
    if (x >= a && x <= b)
        return x;
    else
        return -1;
}
 
// Driver code
var a = 2, b = 10, c = 3;
document.write( getMaxNum(a, b, c));
 
</script>

Output:

Time Complexity: O(1)

Auxiliary Space: O(1)

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