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
