Check if a number is divisible by 47 or not
Given a number N, the task is to check whether the number is divisible by 47 or not.
Examples:
Input: N = 1645
Output: yes
Explanation:
47 * 35 = 1645
Input: N = 4606
Output: yes
Explanation:
47 * 98 = 4606
Approach: The divisibility test of 47 is:
- Extract the last digit.
- Subtract 14 * 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 47, or it is 0, then the original number is also divisible by 47.
For example:
If N = 59173
Step 1:
N = 59173
Last digit = 3
Remaining number = 5917
Subtracting 14 times last digit
Resultant number = 5917 - 14*3 = 5875
Step 2:
N = 5875
Last digit = 5
Remaining number = 587
Subtracting 14 times last digit
Resultant number = 587 - 14*5 = 517
Step 3:
N = 517
Last digit = 7
Remaining number = 51
Subtracting 14 times last digit
Resultant number = 51 - 14*7 = -47
Step 4:
N = -47
Since N is a two-digit number,
and -47 is divisible by 47
Therefore N = 59173 is also divisible by 47
Below is the implementation of the above approach:
C++
#include<bits/stdc++.h>
#include<stdlib.h>
using namespace std;
bool isDivisible( int n)
{
int d;
while (n / 100)
{
d = n % 10;
n /= 10;
n = abs (n-(d * 14));
}
return (n % 47 == 0) ;
}
int main() {
int N = 59173;
if (isDivisible(N))
cout<< "Yes" <<endl ;
else
cout<< "No" <<endl ;
return 0;
}
|
Java
import java.util.*;
class GFG{
static boolean isDivisible( int n)
{
int d;
while ((n / 100 ) > 0 )
{
d = n % 10 ;
n /= 10 ;
n = Math.abs(n - (d * 14 ));
}
return (n % 47 == 0 ) ;
}
public static void main(String[] args) {
int N = 59173 ;
if (isDivisible(N))
System.out.print( "Yes" ) ;
else
System.out.print( "No" );
}
}
|
Python 3
def isDivisible(n) :
while n / / 100 :
d = n % 10
n / / = 10
n = abs (n - (d * 14 ))
return (n % 47 = = 0 )
if __name__ = = "__main__" :
n = 59173
if (isDivisible(n)) :
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
class GFG
{
static bool isDivisible( int n)
{
int d;
while (n / 100 > 0)
{
d = n % 10;
n /= 10;
n = Math.Abs(n - (d * 14));
}
return (n % 47 == 0);
}
public static void Main()
{
int N = 59173;
if (isDivisible(N))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
}
|
Javascript
<script>
function isDivisible(n)
{
let d;
while (Math.floor(n / 100) > 0)
{
d = n % 10;
n = Math.floor(n / 10);
n = Math.abs(n - (d * 14));
}
return (n % 47 == 0) ;
}
let N = 59173;
if (isDivisible(N) != 0)
document.write( "Yes" ) ;
else
document.write( "No" );
</script>
|
PHP
<?php
function isDivisible( $n )
{
while (( $n / 100) <=0)
{
$d = $n % 10;
$n = ( $n /10);
$n = Math. abs ( $n - ( $d * 14));
}
return ( $n % 47 == 0) ;
}
$N = 1645;
if (isDivisible( $N )) {
echo ( "Yes" ) ;
}
else {
echo ( "No" );
}
?>
|
Time Complexity: O(log10N)
Auxiliary Space: O(1)
Last Updated :
22 Jul, 2022
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