$\backslash$table$[[$Op$-$code,Operand,Description$],[1,$RXY$,\backslash$table$[[$LOAD the register $R$ with the bit pattern found in the memory cell whose address is $xY.],[$Example: $14A3$ would cause the contents of the memory cell located at address $A3$ to beplaced in register $4.]]],[2,$RXY$,$Example: $20A3$ would cause the value $A3$ to be placed in register $0.],[3,$RXY$,$STORE the bit pattern found in register R in the memory cell whose address is XY$.],[4,$ORS,MOVE the bit pattern found in register $R$ to register $S.],[$B$,$RXY$,\backslash$table$[[$JUMP to the instruction located in the memory cell at address XY if the bit pattern in register $R$ is equal to the$],[$bit pattern in register number $0.$ Otherwise, continue with the normal sequence of execution. $($The jump is$],[$implemented by copying XY into the program counter during the execute phase.$)],[$Example: B$43$C would first compare the contents of register $4$ with the contents of register $0.$ If the two were$],[$equal$,$ the pattern $3C$ would be placed in the program counter so that the next instruction executed would be$],[$the one located at that memory address. Otherwise, nothing would be done and program execution would$],[$continue in its normal sequence.$]]],[$c$,000,\backslash$table$[[$HALT execution.$],[$Example: CO$00$ would cause program execution to stop.$]]]]$

When answering the following questions, assume that the machine starts with its program counter equal to $20.$

$\backslash$table$[[$Address$,$Content$],[20,12],[21,23],[22,32],[23,30],[24,C0$