Given a number N, the task is to check whether the number is divisible by 31 or not.
Examples:
Input: N = 1922
Output: Yes
Explanation:
31 * 62 = 1922
Input: N = 2722400
Output: No
Approach: The divisibility test of 31 is:
- Extract the last digit.
- Subtract 3 * last digit from the remaining number obtained after removing the last digit.
- Repeat the above steps until a two-digit number, or zero, is obtained.
- If the two-digit number is divisible by 31, or it is 0, then the original number is also divisible by 31.
For example:
If N = 49507 Step 1: N = 49507 Last digit = 7 Remaining number = 4950 Subtracting 3 times last digit Resultant number = 4950 - 3*7 = 4929 Step 2: N = 4929 Last digit = 9 Remaining number = 492 Subtracting 3 times last digit Resultant number = 492 - 3*9 = 465 Step 3: N = 465 Last digit = 5 Remaining number = 46 Subtracting 3 times last digit Resultant number = 46 - 3*5 = 31 Step 4: N = 31 Since N is a two-digit number, and 31 is divisible by 31 Therefore N = 49507 is also divisible by 31
Below is the implementation of the above approach:
C++
// C++ program to check whether a number // is divisible by 31 or not #include<bits/stdc++.h> #include<stdlib.h> using namespace std;
// Function to check if the number is divisible by 31 or not bool isDivisible( int n)
{ int d;
// While there are at least two digits
while (n / 100)
{
// Extracting the last
d = n % 10;
// Truncating the number
n /= 10;
// Subtracting three times the last
// digit to the remaining number
n = abs (n-(d * 3));
}
// Finally return if the two-digit
// number is divisible by 31 or not
return (n % 31 == 0) ;
} // Driver Code int main()
{ int N = 1922;
if (isDivisible(N))
cout<< "Yes" <<endl ;
else
cout<< "No" <<endl ;
return 0;
} // This code is contributed by ANKITKUMAR34 |
Java
// Java program to check whether a number // is divisible by 31 or not import java.util.*;
class GFG{
// Function to check if the number is divisible by 31 or not static boolean isDivisible( int n)
{ int d;
// While there are at least two digits
while ((n / 100 ) > 0 )
{
// Extracting the last
d = n % 10 ;
// Truncating the number
n /= 10 ;
// Subtracting three times the last
// digit to the remaining number
n = Math.abs(n - (d * 3 ));
}
// Finally return if the two-digit
// number is divisible by 31 or not
return (n % 31 == 0 ) ;
} // Driver Code public static void main(String[] args)
{ int N = 1922 ;
if (isDivisible(N))
System.out.print( "Yes" );
else
System.out.print( "No" );
} } // This code is contributed by PrinciRaj1992 |
Python 3
# Python program to check whether a number # is divisible by 31 or not # Function to check if the number is # divisible by 31 or not def isDivisible(n) :
# While there are at least two digits
while n / / 100 :
# Extracting the last
d = n % 10
# Truncating the number
n / / = 10
# Subtracting three times the last
# digit to the remaining number
n = abs (n - (d * 3 ))
# Finally return if the two-digit
# number is divisible by 31 or not
return (n % 31 = = 0 )
# Driver Code if __name__ = = "__main__" :
n = 1922
if (isDivisible(n)) :
print ( "Yes" )
else :
print ( "No" )
|
C#
// C# program to check whether a number // is divisible by 31 or not using System;
class GFG{
// Function to check if the number is divisible by 31 or not static bool isDivisible( int n)
{ int d;
// While there are at least two digits
while ((n / 100) > 0)
{
// Extracting the last
d = n % 10;
// Truncating the number
n /= 10;
// Subtracting three times the last
// digit to the remaining number
n = Math.Abs(n - (d * 3));
}
// Finally return if the two-digit
// number is divisible by 31 or not
return (n % 31 == 0) ;
} // Driver Code public static void Main(String[] args)
{ int N = 1922;
if (isDivisible(N))
Console.Write( "Yes" );
else
Console.Write( "No" );
} } // This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript program to check whether a number // is divisible by 31 or not // Function to check if the number is divisible by 31 or not function isDivisible(n)
{ let d;
// While there are at least two digits
while (Math.floor(n / 100) > 0)
{
// Extracting the last
d = n % 10;
// Truncating the number
n = Math.floor(n / 10);
// Subtracting three times the last
// digit to the remaining number
n = Math.abs(n - (d * 3));
}
// Finally return if the two-digit
// number is divisible by 31 or not
return (n % 31 == 0) ;
} // Driver Code let N = 1922;
if (isDivisible(N) != 0)
document.write( "Yes" ) ;
else
document.write( "No" );
// This code is contributed by sanjoy_62. </script> |
Output:
Yes
Time Complexity: O(log10N)
Auxiliary Space: O(1)