CareerCup

I think this may work

import java.io.*;

class Main
{
	public static void main(String ar[])
	{
		int n=43;
		double temp = 1;
		while( (temp%n) != 0)
		{
			temp = (temp*10) +1;
			System.out.println(temp);
		}
		System.out.println(temp);
	}

}

This will work

int main(void)
{
     int arr[100000];
     for(int i=0;i<100000;i++)
     arr[i]=1;
     
     int i,n,num=0;
     cin>>n;
     
     for( i=0;i<1000000;i++)
     {
                num=(num*10+arr[i])%n;
                if(num==0)
                break;
     }
     
     for(;i>=0;i--)
     cout<<arr[i];
     getch();
     
}

Please comment on my solution, I think it works, Thanks!

int test1s(int n)
{
    int num1s = 0;
    while(n%10==1)
    {
        n = n/10;
        num1s++;
    }
if( n == 0 )
    return num1s;
else
    return 0;
}

string minmul3(int n)
{
string err = "err";
if (mod(n+10,10)!=3)
return err;

//we are solving the problem by iterative find the digit of b,which satisfies n*b=111....11.
//pre recomputed map[i] = s; st: mod(s*3+i,10) = 1;
int map[] = {7,0,3,6,9,2,5,8,1,4};

int res =0;
int count = 0;
while (test1s(res)==0)
{
	count++;
	int b = map[res%10];
	res = res/10 + b*n;
}
count = count + test1s(res);
string result;
result.resize(count+1);
memset(result.c_str(),'1',count);
result.c_str()[count+1] = 0;
return result;
}

#include<iostream.h>
#include<conio.h>
#include<math.h>
int main()
{
clrscr();
int a,i,b,count=1;
cout<<"Enter a";
cin>>a;
for(i=1;;i+=b)
{
count++;
b=pow(10,count);
if(a%10!=3)
{cout<<"\n wrong output";
getch();
return 0;
}
else
{
if(i%10==1)

{
if(i%a==0)
{
cout<<"\n Ans is "<<i<<" is divided by "<<a ;
getch();
return 0;
}
}
}
}
getch();
}

I cant understand the question..could someone plz elaborate?

#include"stdio.h"
#define SIZE 100000

int modulo(int num[],int n)
{
int pre_ten=1;
int pre_mod=0;
int i=0,count=0;
while(i<SIZE)
{
count++; i++;
pre_mod=(pre_mod + pre_ten)%n ;
if(pre_mod==0) break ;
pre_ten=(pre_ten*10)%n ;
}
if(i==SIZE) return -1;
return count ;
}

// I assume that answer will come within 100000 digits

int main()
{
int num[SIZE] ;
int i;
for(i=0;i<SIZE;i++)
num[i]=1;
int n;
scanf("%d",&n);
int c=modulo(num,n) ;
printf("%d\n",c);
// c is total 1's in answer
// so we have answer string.
return 0;
}

start with no. of 1's as 0 in answer. Then dynamically increase no. of 1's in answer until we get the answer. We will get the ans when answer modulo n becomes zero. use this : (a+b)%n= (a%n + b%n)%n.
let at some time ans is 1111 (it mean we have 1111%n), then
11111%n= (10000 + 1111)%n = (10000%n + 1111%n)%n.
also 10000%n= (10*(1000%n))%n ( use of (a*b)%n= (a*(b%n))%n )

#include"stdio.h"
#define SIZE 100000

int modulo(int num[],int n)
{
	int pre_ten=1;
	int pre_mod=0;
	int i=0,count=0;
	while(i<SIZE)
	{
		count++; i++;
		pre_mod=(pre_mod + pre_ten)%n ;
		if(pre_mod==0) break ;
		pre_ten=(pre_ten*10)%n ;
	}
	if(i==SIZE) return -1;
	return count ;
}

// I assume that answer will come within 100000 digits

int main()
{
	int num[SIZE] ;
	int i;
	for(i=0;i<SIZE;i++)
	num[i]=1;
	int n;
	scanf("%d",&n);
	int c=modulo(num,n) ;
	printf("%d\n",c);
	// c is total 1's in answer
	// so we have answer string.
	return 0;
}

# In Python 
    n = int(sys.argv[1])  # get the integer 
    count = 1
    base = 1
    while base < n:
            base *= 10
            base +=1
            count += 1
    reminder  = base % n
    while reminder > 0:
            reminder *= 10
            reminder += 1
            reminder = reminder % n
            count+=1
    string = ''
    for i in range(0, count)
            string += '1'
    print string

one point
You can get all 1's for n only if n = (2 pow k) -1. correct me if i am wrong.If it is true shouldn't we only try for ((2 pow k )-1)%a ==0.

public static void testFunction(int num) {
int count = 1, rm = 1;
while (rm > 0) {
rm = (rm * 10 + 1) % num;
count++;
}
while (count-- > 0)
System.out.print("1");
}

#include<iostream>
using namespace std;
int main(){
cout<<"enter number";
unsigned int n;
cin>>n;
int n1=((1<<(n*2))-1);
printf("%x is divisible by %d",n1,n); //there is convert the n1 into binary
//the get the no. of 1's which is multile of n

return 0;
}