please, Help me to convert String to Binary

Hi every one,
I am new student in java programming, so I need your help

I want to takes a function as its input a string of decimal integers, each character in the string can be thought of as a decimal digit. The digit should be converted to 3-bit binary string an packed into an int.

I did the following code, but there is problem in converting 5 and 6 also I want the output like this

1234567 = 001 010 011 100 101 110 111

class Bin
{
public static void main(String []args)
{
String N=(args[0]);
for(int i=0;i<N.length();i++)
{
String n=N.substring(i,i+1);
int a=Integer.parseInt(n);
int y=1,b;
String z="";

for(a=a;a!=0; )
{
z=z+y;
y=a&1;
a=a>>1;

}
b=Integer.parseInt(z);
System.out.print(" "+b+" ");

}
System.out.println();
}
}

Thanks,

import java.io.*;

public class BinaryCalculator {
  private static final int maxBytes = 3;
  public static void main(String[] args) {
    // we use a reader to read the inputs from the keyboard
    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    do {
      try {
        System.out.print("Type the number to parse: ");
        // if the number cannot be parsed, an exception will be thrown and is catched further down
        int number = Integer.parseInt(in.readLine());
        int currentBit;
        String result = "";
        // this is where we calculate the number in the same manner as we would do by hand.
        // since the maximum number of bytes are known, we can simply run through a for-loop, 
        // or we would have to make further calculations to figure out the size of the most-significant-bit
        for (int i = maxBytes*8; i >= 0; i--) {
          currentBit = 1 << i;
          if (number >= currentBit) {
            result += 1;
            number -= currentBit;
          }
          else {
            result += 0;
          }
        }
        // and then printing the result to the screen
        System.out.println(result);
      }
      // if we tried to parse a string that was not pure numbers, this will be thrown. 
      catch (NumberFormatException e) {
        // the program exits
        System.exit(0);
      }
      // if the reader threw an exception, something is wrong. alas, we print the stack-trace and exit
      catch (IOException e) {
        e.printStackTrace();
        System.exit(1);
      }
    }
    while (true);
  }
}

  for (x=0;x<3;++x) {
     if ((a & 0x40)!=0)
  	  print '1'
  	  a<<=1
   else
  	  print '0'
    }
   print a space