Repeated Unit Divisibility
Last Updated :
17 Jul, 2022
A number which consists entirely repeated number of times one is said to be Repeated Unit. We shall define R(k) of length k repeated unit number. For example, R(6) = 111111.
For a given number n, a positive integer and GCD(n, 10) = 1, there exists a value k such that R(k) is divisible by n. Now let A(n) be the least such value of k i.e. A(n) = k.
Thus, we have to find for the given value of n the least such value of A(n) for which k times repeated one is divided by n.
Examples:
Input : n = 7
Output : A(7) = 6
A(7) = 6 means 6 times one i.e. (111111) is divided by 7.
Input : n = 41
Output : A(41) = 5
A(41) = 5 means 5 times one i.e. (11111) is divided by 41.
First of all if the given number should be coprime with 10 otherwise return 0.
If the given number is coprime with 10 then we have to find smallest number k such that R(k) = 0 mod n.
Consider the repeated units R(1), R(2), R(3), R(4) and so on. For each repeated unit R(j) suppose when calculating remainder of R(j) divided by n. There are n conceivable remainders. We get that R(i), R(j) where i < j have same remainder when divided by n. It follows that R(j) – R(i) is divided by n. This difference is repeated unit multiplied by power of 10. But, as we know that 10 and n are relatively prime, n divides R(k).
C++
#include <bits/stdc++.h>
using namespace std;
int repUnitValue( int n)
{
if (n % 2 == 0 || n % 5 == 0)
return 0;
int rem = 1;
int power = 1;
int k = 1;
while (rem % n != 0)
{
k++;
power = power * 10 % n;
rem = (rem + power) % n;
}
return k;
}
int main()
{
int n = 13;
cout << repUnitValue(n);
return 0;
}
|
Java
import java.util.*;
class GFG {
public static int repUnitValue( int n)
{
if (n % 2 == 0 || n % 5 == 0 )
return 0 ;
int rem = 1 ;
int power = 1 ;
int k = 1 ;
while (rem % n != 0 )
{
k++;
power = power * 10 % n;
rem = (rem + power) % n;
}
return k;
}
public static void main(String[] args)
{
int n = 13 ;
System.out.println(repUnitValue(n));
}
}
|
Python3
def repUnitValue(n):
if (n % 2 = = 0 or n % 5 = = 0 ):
return 0
rem = 1
power = 1
k = 1
while (rem % n ! = 0 ):
k + = 1
power = power * 10 % n
rem = (rem + power) % n
return k
n = 13
print (repUnitValue(n))
|
C#
using System;
public class GFG {
public static int repUnitValue( int n)
{
if (n % 2 == 0 || n % 5 == 0)
return 0;
int rem = 1;
int power = 1;
int k = 1;
while (rem % n != 0)
{
k++;
power = power * 10 % n;
rem = (rem + power) % n;
}
return k;
}
public static void Main()
{
int n = 13;
Console.Write(repUnitValue(n));
}
}
|
PHP
<?php
function repUnitValue( $n )
{
if ( $n % 2 == 0 || $n % 5 == 0)
return 0;
$rem = 1;
$power = 1;
$k = 1;
while ( $rem % $n != 0)
{
$k ++;
$power = $power * 10 % $n ;
$rem = ( $rem + $power ) % $n ;
}
return $k ;
}
$n = 13;
echo repUnitValue( $n );
?>
|
Javascript
<script>
function repUnitValue(n)
{
if (n % 2 == 0 || n % 5 == 0)
return 0;
let rem = 1;
let power = 1;
let k = 1;
while (rem % n != 0)
{
k++;
power = power * 10 % n;
rem = (rem + power) % n;
}
return k;
}
let n = 13;
document.write(repUnitValue(n));
</script>
|
Time complexity: O(n)
Auxiliary Space: O(1)
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