In the familiar base 10 system the smallest rightmost digit

Question

The Binary Number System

In the familiar base 10 system, the smallest (rightmost) digit of a number is the ones place (10⁰).  The next digit is in the tens place (10¹), and the next one is in the hundreds place (10²) and so forth.  Thus the number 122 is really (1 * 10²) + (2 * 10¹) + (2 * 10⁰).  Each digit is worth a power of 10 more than the digit to the right of it in the base 10 system.

Similarly, in the binary number system, the smallest (rightmost) digit of a number is the ones place (2⁰).  The next digit is in the twos place (2¹), and the next one is in the fours place (2²) and so forth.  Thus the number 101 is really (1 * 2²) + (0* 2¹) + (1 * 2⁰) = 5.  Each digit is worth a power of 2 more than the digit to the right of it in the base 2 system.

Write a program to display all of the binary digits of a number.  You program should continue

to display binary digits of numbers until the user wishes to stop.

Hint:

To find a binary digit, print the remainder number % 2, then replace the number with number/2.  Repeat this process until the number is 0.

SAMPLE OUTPUT:

Enter a number for binary conversion=>13
Binary Number
1
0
1
1

Another number to enter =? , type 1 for yes and any other number for no=>1
Enter a number for binary conversion=>10
Binary Number
0
1
0
1
Another number to enter =? , type 1 for yes and any other number for no=>2