>>BITXOR(X,N)=Y
>>BITXOR(Y,N)=X
>>
>>how do we find the value of 'N'
>>
>>Thanks to anyone who can illuminate
>>
>Harry,
>
>Sergey's has already given you the correct syntax. Just to add a bit more information, bits in the same position result in the following

A   B   Result
>0   0     0
>1   0     1
>0   1     1
>1   1     0

To sum it up, if bits in the same position match, the return value is that the bit in the return value is not set (it's cleared). If they do not match, the bit is set.

Further to George & Sergey,

If BITXOR(X,N)=Y
then all variations on X,N,Y are true

BITXOR(N,X)=Y
BITXOR(Y,N)=X
BITXOR(N,Y)=X
BITXOR(Y,X)=N
BITXOR(X,Y)=N