every odd no. has "1" in its last bit and every even no has "0" (in binary)
eg-
2=0010 ,3=0011
4=0100 ,5=0101....and so on
so just logically "AND" the no with "01H" using instruction [HAVING THAT NO. IN ACCUMULATOR (A)]
ANI 01H
so if the no. is even =>
A=0 (and zero flag is set i.e. "1"
but if the no is odd then => A=1
means zero flag is reset i.e. "0"
so using instruction "JZ"
you can find if the no. if even or odd. :)
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