Given a number, the task is to check if number is divisible by 20. The input number may be large and it may not be possible to store long long int and it may be very large number then we use the string.
Examples:
Input : 7575680 Output : Yes Input : 987985865687690 Output : No
A number is divisible by 20 if it is divisible by 5 and 4. We can check if a number is divisible by 4 by checking if last two digits are divisible by 4. We can check for divisibility by 5 by checking last digit. Also, if the last digit of a number is zero and the second last digit is a multiple of 2 then the number is divisible by 20.
// CPP program to check if a large number // is divisible by 20. #include <iostream> using namespace std;
bool divisibleBy20(string num)
{ // Get number with last two digits
int lastTwoDigits = stoi(num.substr(num.length() - 2,
num.length() - 1));
// Check if the number formed by last two
// digits is divisible by 5 and 4.
return ((lastTwoDigits % 5 == 0) &&
(lastTwoDigits % 4 == 0));
} int main()
{ string num = "63284689320" ;
if (divisibleBy20(num))
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
} |
// Java program to check if a large n // number is divisible by 20. import java.io.*;
class GFG {
static Boolean divisibleBy20(String num)
{
// Get number with last two digits
int lastTwoDigits = Integer.parseInt(num.substring(num.length() - 2 ,
num.length() ));
// Check if the number formed by last two
// digits is divisible by 5 and 4.
return ((lastTwoDigits % 5 == 0 ) &&
(lastTwoDigits % 4 == 0 ));
}
// Driver Program
public static void main (String[] args)
{
String num = "63284689320" ;
if (divisibleBy20(num) == true )
System.out.println( "Yes" );
else
System.out.println( "No" );
}
} // This code is contributed by Gitanjali. |
# Python3 program to check if a large # number is divisible by 20. import math
def divisibleBy20(num):
# Get number with last two digits
lastTwoDigits = int (num[ - 2 :])
# Check if the number formed by last two
# digits is divisible by 5 and 4.
return ((lastTwoDigits % 5 = = 0 and
lastTwoDigits % 4 = = 0 ))
# driver code num = "63284689320"
if (divisibleBy20(num) = = True ):
print ( "Yes" )
else :
print ( "No" )
# This code is contributed by Gitanjali. |
// C# program to check if a large // 'n' number is divisible by 20. using System;
using System.Text;
class GFG
{ static bool divisibleBy20(String num)
{ // Get number with last two digits
int lastTwoDigits = Int32.Parse(num.Substring(2));
// Check if the number formed
// by last two digits is
// divisible by 5 and 4.
return ((lastTwoDigits % 5 == 0) &&
(lastTwoDigits % 4 == 0));
} // Driver Code static public void Main ()
{ String num = "63284689320" ;
if (divisibleBy20(num) == true )
Console.Write( "Yes" );
else
Console.Write( "No" );
} } // This code is contributed by Raj |
<?php // PHP program to check // if a large number is // divisible by 20. function divisibleBy20( $num )
{ // Get number with
// last two digits
$lastTwoDigits = intval ( substr ( $num ,
( strlen ( $num ) - 2), 2));
// Check if the number
// formed by last two
// digits is divisible
// by 5 and 4.
return (( $lastTwoDigits % 5 == 0) &&
( $lastTwoDigits % 4 == 0));
} // Driver Code $num = "63284689320" ;
if (divisibleBy20( $num ))
echo "Yes" ;
else echo "No" ;
// This code is contributed by mits. ?> |
<script> // Javascript program to check // if a large number is // divisible by 20. function divisibleBy20(num)
{ // Get number with
// last two digits
let lastTwoDigits = parseInt(num.slice(-2, num.length))
console.log(num.slice(-2, 1))
// Check if the number
// formed by last two
// digits is divisible
// by 5 and 4.
return ((lastTwoDigits % 5 == 0) &&
(lastTwoDigits % 4 == 0))
} // Driver Code let num = "63284689320" ;
if (divisibleBy20(num))
document.write( "Yes" );
else document.write( "No" );
// This code is contributed by _saurabh_jaiswal. </script> |
Yes
Time complexity: O(1)
Auxiliary space: O(1)
Method: Checking given number is divisible by 20 or not by using modulo division operator “%”.
#include <iostream> using namespace std;
int main() {
// input number
long int num = 987985865687690;
// finding given number is divisible by 20 or not
if (num % 20 == 0) {
cout<< "Yes" ;
}
else {
cout<< "No" ;
}
return 0;
} // This code is contributed by laxmigangarajula03 |
// Java code for the above approach // To check whether the given number is divisible by 20 or // not import java.math.BigInteger;
class GFG {
public static void main(String[] args)
{
// input number
BigInteger num = new BigInteger( "987985865687690" );
// finding given number is divisible by 20 or not
if (num.mod( new BigInteger( "20" ))
.equals( new BigInteger( "0" ))) {
System.out.println( "Yes" );
}
else {
System.out.println( "No" );
}
}
} // This code is contributed by phasing17 |
# Python code # To check whether the given number is divisible by 20 or not #input n = 987985865687690
# the above input can also be given as n=input() -> taking input from user # finding given number is divisible by 20 or not if int (n) % 20 = = 0 :
print ( "Yes" )
else :
print ( "No" )
# this code is contributed by gangarajula laxmi
|
// C# code for the above approach // To check whether the given number is divisible by 20 or // not using System;
public class GFG {
static public void Main()
{
// input number
long num = 987985865687690;
// finding given number is divisible by 20 or not
if (num % 20 == 0) {
Console.Write( "Yes" );
}
else {
Console.Write( "No" );
}
}
} // This code is contributed by laxmigangarajula03 |
<script> // JavaScript code for the above approach
// To check whether the given number is divisible by 20 or not
//input
var n = 987985865687690
// finding given number is divisible by 20 or not
if (n % 20 == 0)
document.write( "Yes" )
else
document.write( "No" )
// This code is contributed by Potta Lokesh
</script>
|
<?php // PHP program to check // if a large number is // divisible by 20. // Driver Code
// input number
$num = 987985865687690;
// finding given number is divisible by 20 or not if ( ( $num ) % 20 == 0)
echo "Yes" ;
else echo "No" ;
// This code is contributed by satwik4409. ?> |
No
Time complexity: O(1) it is performing constant operations
Auxiliary space: O(1)