Unit 2 Practice Test Coding Problem 5 (External resource)

Does anyone know how to write this code to print out the binary.. specifically using floor division and modulus ... im very lost

\#You may modify the lines of code above, but don't move them! \#When you submit your code, we'll change these lines to \#assign different values to the variables. \# \#The number above is given in base 10. Let's convert it to \#base 2 and print it in binary. For example, 215 can be written \#in binary as 11010111. \# \#Each digit of that string corresponds to a power of 2 . The far \#left digit represents 128 , then 64 , then 32 , then 16 , then 8 , \#then 4, then 2, and then finally 1 at the far right. \# \#To convert the number to binary, first check to see if it is \#greater than or equal to 128 . If it is, your first digit is 1. \#If not, your first digit is 0 . If the number was greater than \#128, subtract 128 (because the 128 is captured by the first \#digit of the string). \# \#Then, repeat that process for 64,32,16,8,4,2, and 1. \# \#For example: \# \#215 is >= 128:1 #87 is >=64:11 \#23 is not >=32:110 #23 is >=16:1101 #7 is not >= 8:11010 \#7 is >=4:110101 #3 is >=2:1101011 \#1 is >= 1: 11010111 \# \#Note that although we use 'if' a lot to describe this problem, \#this can be done entirely with floor division and modulus. \#Remember, if you convert a boolean to an integer, True becomes #1 and False becomes 0 . 9 \#Note that we always work with binary in 8-bit chunks : the 10 \#number 7 would be 00000111 , not just 111 . That s because inside 11 \#the computer, 8 ss and 0 m make a byte, which is the smallest 12 \#practical unit of storage (rarely are bits used outside 8 -bit 3 \#bytes). 4 \# \# 15 \#Print the string that results from this conversion. 17 18 \#Add your code here! 9 \#