has only the prime divisors and .
Every divisor of is therefore of the form ,
where and are elements of {0,1,2,3}. Since there
are choices for and choices for , there are
divisors of . That isn't necessary to know. But
in the product of all divisors of ,
or , and or occur exactly times each.
Since and both occur times each in the product
of all divisors, the product of the divisors must be
, choice a)
Edwin