Answer:
(A) Assume m is an odd integer.
Therefore m is of the m=2n-1 type where n is any number.
We have m + 6 = 2n-1 + 6 = 2n+5=2n+4 + 1 = 2(n+2) + 1, in which n + 2 is an integer, adding 6 to both ends.
Because m+6 is of the 2x + 1 type, where x = n + 2; then m + 6 is an odd integer too.
(B) Provided that mn is an integer also. For some integer y and m, n are integers, this means mn is of the form mn = 2y.
And y = mn/2.
Therefore either m is divided by 2 or n is divided by 2 since y, m, n are all integers. To put it another way, either m is a multiple of 2 or n is a multiple of 2, which means that m is even or n is even.
