求一个数这个数只能被它自己和它自己的平方根和1整除
A:121 is the product of only 11 and 11 ,or 1 and 121.Since 11 is the square root of 121,then 121 has only three positive integer factors,itself 121 ,its square root 11 ,and 1
B:100 is the product of 100 and 1,20 and 5,2 and 50,4 and 25,etc.then 100 has more than 3 positive integer factor
c:81=9*9,81*1,3*27,where 3*27 is not itself nor its square root nor 1
d:64=8*8,2*32,where 2*32 is not itself not its square nor 1
e:33=1*33,3*11,and so it has only 2 positive integer factors.
then answer is A.