Def sum_divisors(n): sum =1 # Return the sum of all divisors of n, not including n x=int(n**0.5) for i in range(2,(x//1)+1): if n%i==0: sum=sum+i if n%i==0 and n/i!=i: sum=sum+(n/i) return sum print(sum_divisors(0)) # 0 print(sum_divisors(3)) # Should sum of 1 # 1 print(sum_divisors(36)) # Should sum of 1+2+3+4+6+9+12+18 # 55 print(sum_divisors(102)) # Should be sum of 2+3+6+17+34+51 # 114

Write a new lc-3 trap subroutine (i.e. a subroutine that will be invoked via the trap instruction) that will receive a numeric digit entered at the keyboard (i.e. an ascii character), echo it to the screen, and return in r0 the corresponding numeric value: so if the user types the digit '7', the character '7' will appear on the screen, but the value returned in r0 will be b0000 0000 0000 0111 (#7) you may not use any trap calls in your code - you must implement the "polling" code that interrogates the keyboard status and data registers. ; getnum_tsr ; a subroutine for obtaining a numeric value ; given ascii numeric digit input to keyboard. ; the numeric digit is echoed to the console (e.g. '7' = b0000 0000 0011 0111), ; but the value returned in r0 is the actual numeric value ; corresponding to the digit (e.g. b0000 0000 0000 0111 =