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++
// CPP program to find least value of// k for which R(k) is divisible by n#include <bits/stdc++.h>using namespace std;// To find least value of kint repUnitValue(int n){ // To check n is coprime or not if (n % 2 == 0 || n % 5 == 0) return 0; // to store R(k) mod n and 10^k // mod n value int rem = 1; int power = 1; int k = 1; while (rem % n != 0) { k++; power = power * 10 % n; rem = (rem + power) % n; } return k;}// Driver codeint main(){ int n = 13; cout << repUnitValue(n); return 0;} |
Java
// Java program to find least value of// k for which R(k) is divisible by nimport java.util.*;class GFG { // To find least value of k public static int repUnitValue(int n) { // To check n is coprime or not if (n % 2 == 0 || n % 5 == 0) return 0; // to store the R(k) mod n and // 10^k mod n value int rem = 1; int power = 1; int k = 1; while (rem % n != 0) { k++; power = power * 10 % n; rem = (rem + power) % n; } return k; } // Driver code public static void main(String[] args) { int n = 13; System.out.println(repUnitValue(n)); }} |
Python3
# python program to find least value of# k for which R(k) is divisible by n# To find least value of kdef repUnitValue(n): # To check n is coprime or not if (n % 2 == 0 or n % 5 == 0): return 0 # to store R(k) mod n and 10^k # mod n value rem = 1 power = 1 k = 1 while (rem % n != 0): k += 1 power = power * 10 % n rem = (rem + power) % n return k # Driver coden = 13print(repUnitValue(n))# This code is contributed by Sam007. |
C#
// C# program to find least value of// k for which R(k) is divisible by nusing System;public class GFG { // To find least value of k public static int repUnitValue(int n) { // To check n is coprime or not if (n % 2 == 0 || n % 5 == 0) return 0; // to store the R(k) mod n and // 10^k mod n value 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)); }}// This code is contributed by Sam007. |
PHP
<?php// PHP program to find least value of// k for which R(k) is divisible by nfunction repUnitValue($n){ // To check n is coprime or not if ($n % 2 == 0 || $n % 5 == 0) return 0; // to store R(k) mod n // and 10^k mod n value $rem = 1; $power = 1; $k = 1; while ($rem % $n != 0) { $k++; $power = $power * 10 % $n; $rem = ($rem + $power) % $n; } return $k;} // Driver Code $n = 13; echo repUnitValue($n); // This code is contributed by aj_36?> |
Javascript
<script>// java script program to find least value of// k for which R(k) is divisible by nfunction repUnitValue(n){ // To check n is coprime or not if (n % 2 == 0 || n % 5 == 0) return 0; // to store R(k) mod n // and 10^k mod n value let rem = 1; let power = 1; let k = 1; while (rem % n != 0) { k++; power = power * 10 % n; rem = (rem + power) % n; } return k;} // Driver Code let n = 13; document.write(repUnitValue(n)); // This code is contributed by sravan kumar.</script> |
6
Time complexity: O(n)
Auxiliary Space: O(1)
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