Given a positive integer N whose unit’s digit is 3. Find the number of 1s in the smallest repunit which is divisible by the given number N. Every number whose unit’s digit is 3 has a repunit as its multiple. A repunit is a number which has only ones. It is of the form (10n – 1)/9.
Examples:
Input: 3
Output: 3
As 3 divides 111 which is the smallest repunit
multiple of the number. So the number of ones in 111 is 3.
Input: 13
Output: 6
The repunits are 1, 11, 111, 1111, …. the next repunit to x will always be x*10+1. If the remainder left by x repunit is r then remainder left by the next repunit will always be (r*10+1)%n. Since the repunit can be very large, there is no need to find the repunit number. Simply counting the number of ones will give us the answer.
So, find out the remainders of all repunit numbers until the remainder becomes 0. Once it does, then the count of iterations done to make remainder 0 will be the number of 1’s.
Below is the implementation of above approach :
C++
// CPP program to print the number of 1s in// smallest repunit multiple of the number.#include <bits/stdc++.h>using namespace std;// Function to find number of 1s in// smallest repunit multiple of the numberint countOnes(int n){ // to store number of 1s in smallest // repunit multiple of the number. int count = 1; // initialize rem with 1 int rem = 1; // run loop until rem becomes zero while (rem != 0) { // rem*10 + 1 here represents // the repunit modulo n rem = (rem * 10 + 1) % n; count++; } // when remainder becomes 0 // return count return count;}// Driver Codeint main(){ int n = 13; // Calling function cout << countOnes(n);} |
Java
// Java program to print the // number of 1s in smallest // repunit multiple of the number.import java.io.*;class GFG {// Function to find number // of 1s in smallest repunit// multiple of the numberstatic int countOnes(int n){ // to store number of 1s // in smallest repunit // multiple of the number. int count = 1; // initialize rem with 1 int rem = 1; // run loop until // rem becomes zero while (rem != 0) { // rem*10 + 1 here // represents the // repunit modulo n rem = (rem * 10 + 1) % n; count++; } // when remainder becomes // 0 return count return count;}// Driver Codepublic static void main (String[] args){int n = 13;// Calling functionSystem.out.println(countOnes(n));}}// This code is contributed by akt_mit |
Python3
# Python3 program to print the # number of 1s in smallest# repunit multiple of the number.# Function to find number # of 1s in smallest repunit# multiple of the numberdef countOnes(n): # to store number of 1s # in smallest repunit # multiple of the number. count = 1; # initialize rem with 1 rem = 1; # run loop until # rem becomes zero while (rem != 0): # rem*10 + 1 here represents # the repunit modulo n rem = (rem * 10 + 1) % n; count = count + 1; # when remainder becomes # 0 return count return count;# Driver Coden = 13;# Calling functionprint(countOnes(n)); # This code is contributed by mits |
C#
// C# program to print the // number of 1s in smallest // repunit multiple of the number.using System;class GFG{ // Function to find number // of 1s in smallest repunit// multiple of the numberstatic int countOnes(int n){ // to store number of 1s // in smallest repunit // multiple of the number. int count = 1; // initialize rem with 1 int rem = 1; // run loop until // rem becomes zero while (rem != 0) { // rem*10 + 1 here represents // the repunit modulo n rem = (rem * 10 + 1) % n; count++; } // when remainder becomes 0 // return count return count;}// Driver Codestatic public void Main (){int n = 13;// Calling functionConsole.WriteLine (countOnes(n));}}// This code is contributed by ajit |
PHP
<?php// PHP program to print the // number of 1s in smallest// repunit multiple of the number.// Function to find number // of 1s in smallest repunit// multiple of the numberfunction countOnes($n){ // to store number of 1s // in smallest repunit // multiple of the number. $count = 1; // initialize rem with 1 $rem = 1; // run loop until // rem becomes zero while ($rem != 0) { // rem*10 + 1 here represents // the repunit modulo n $rem = ($rem * 10 + 1) % $n; $count++; } // when remainder becomes // 0 return count return $count;}// Driver Code$n = 13;// Calling functionecho countOnes($n); // This code is contributed by ajit?> |
Javascript
<script> // Javascript program to print the // number of 1s in smallest // repunit multiple of the number. // Function to find number // of 1s in smallest repunit // multiple of the number function countOnes(n) { // to store number of 1s // in smallest repunit // multiple of the number. let count = 1; // initialize rem with 1 let rem = 1; // run loop until // rem becomes zero while (rem != 0) { // rem*10 + 1 here represents // the repunit modulo n rem = (rem * 10 + 1) % n; count++; } // when remainder becomes 0 // return count return count; } let n = 13; // Calling function document.write(countOnes(n)); // This code is contributed by rameshtravel07.</script> |
Output:
6
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