I need help with solving this Binary puzzle (LC-3 Instruction)

A value is stored in R5 before this program runs.

(This is the instruction)

After the following program is executed, the value stored in memory cell x330B is 4. The last instruction is a TRAP instruction which halts execution.

Which of the following possible initial values for R5 will produce this result?

Check all options (1-14) that apply:

1) 1110 1110 1110 1110

2) 1011 1011 1001 1011

3) 1011 1011 1010 1001

4) 1111 1111 0000 1111

5) 0010 1000 1100 0111

6) 1111 0011 1000 0011

7) 1111 1111 1101 1001

8) 1010 1010 0101 0101

9) 1111 0000 0000 0000

10) 1111 1111 1110 0100

11) 1111 1111 1111 1111

12) 1110 1110 1101 0001

13) 1111 0000 1000 0001

14) 0000 1111 0000 0001

address 4 bits 12 bits 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 X32FF 1 0 1 0 0 0 0 1 0 x3300 1 0 1 1 1 1 1 1 1 1 0 0 0 0 0 x3301 0 0 1 1 1 1 1 1 1 0 1 x3302 x3302 0 1 0 1 1 0 0 1 0 1 0 0 1 1 0 x3303 1 e 1 x3304 0 1 e 1 e e e e 1 x3305 0 1 1 1 0 1 0 1 1 0 1 1 0 x3306 1 1 1 1 1 1 1 1 0 1 x3307 0 0 1 0 0 1 1 1 1 1 1 1 1 x3308 0 0 0 1 0 1 1 1 1 1 1 1 0 0 1 x3309 1 1 e e e e e e e e e 1 x330 1 1 1 1 1 0 0 0 0 0 0 1 0 0 1 0 1 address 4 bits 12 bits 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 X32FF 1 0 1 0 0 0 0 1 0 x3300 1 0 1 1 1 1 1 1 1 1 0 0 0 0 0 x3301 0 0 1 1 1 1 1 1 1 0 1 x3302 x3302 0 1 0 1 1 0 0 1 0 1 0 0 1 1 0 x3303 1 e 1 x3304 0 1 e 1 e e e e 1 x3305 0 1 1 1 0 1 0 1 1 0 1 1 0 x3306 1 1 1 1 1 1 1 1 0 1 x3307 0 0 1 0 0 1 1 1 1 1 1 1 1 x3308 0 0 0 1 0 1 1 1 1 1 1 1 0 0 1 x3309 1 1 e e e e e e e e e 1 x330 1 1 1 1 1 0 0 0 0 0 0 1 0 0 1 0 1