Given a number N, the task is to check whether the number is divisible by 71 or not.
Examples:
Input: N = 25411681
Output: yes
Explanation:
71 * 357911 = 25411681Input: N = 5041
Output: yes
Explanation:
71 * 71 = 5041
Approach: The divisibility test of 71 is:
- Extract the last digit.
- Subtract 7 * 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 71, or it is 0, then the original number is also divisible by 71.
For example:
If N = 5041 Step 1: N = 5041 Last digit = 1 Remaining number = 504 Subtracting 7 times last digit Resultant number = 504 - 7*1 = 497 Step 2: N = 497 Last digit = 7 Remaining number = 49 Subtracting 7 times last digit Resultant number = 49 - 7*7 = 0 Step 3: N = 0 Since N is a two-digit number, and 0 is divisible by 71 Therefore N = 5041 is also divisible by 71
Below is the implementation of the above approach:
C++
// C++ program to check whether a number // is divisible by 71 or not #include<bits/stdc++.h> #include<stdlib.h> using namespace std;
// Function to check if the number is divisible by 71 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 seven times the last
// digit to the remaining number
n = abs (n - (d * 7));
}
// Finally return if the two-digit
// number is divisible by 71 or not
return (n % 71 == 0) ;
} // Driver Code int main() {
int N = 5041;
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 71 or not import java.util.*;
class GFG{
// Function to check if the number is divisible by 71 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 seven times the last
// digit to the remaining number
n = Math.abs(n - (d * 7 ));
}
// Finally return if the two-digit
// number is divisible by 71 or not
return (n % 71 == 0 ) ;
}
// Driver Code
public static void main(String args[]){
int N = 5041 ;
if (isDivisible(N))
System.out.println( "Yes" ) ;
else
System.out.println( "No" );
}
} // This code is contributed by AbhiThakur |
Python 3
# Python program to check whether a number # is divisible by 71 or not # Function to check if the number is # divisible by 71 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 seven times the last
# digit to the remaining number
n = abs (n - (d * 7 ))
# Finally return if the two-digit
# number is divisible by 71 or not
return (n % 71 = = 0 )
# Driver Code if __name__ = = "__main__" :
N = 5041
if (isDivisible(N)) :
print ( "Yes" )
else :
print ( "No" )
|
C#
// C# program to check whether a number // is divisible by 71 or not using System;
class GFG
{ // Function to check if the number is divisible by 71 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 fourteen times the last
// digit to the remaining number
n = Math.Abs(n - (d * 7));
}
// Finally return if the two-digit
// number is divisible by 71 or not
return (n % 71 == 0);
} // Driver Code public static void Main()
{ int N = 5041;
if (isDivisible(N))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
} } // This code is contributed by mohit kumar 29. |
Javascript
<script> // Javascript program to check whether a number // is divisible by 71 or not // Function to check if the number is divisible by 71 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 seven times the last
// digit to the remaining number
n = Math.abs(n - (d * 7));
}
// Finally return if the two-digit
// number is divisible by 71 or not
return (n % 71 == 0) ;
} // Driver Code let N = 5041; if (isDivisible(N))
document.write( "Yes" ) ;
else document.write( "No" );
// This code is contributed by patel2127 </script> |
Output:
Yes
Time Complexity: O(log10N)
Auxiliary Space: O(1)