A pack of numbers ...

For having 6 divisors , number has to be of the form

$p^5$ or $p_1 p_2^2$ because then the number of divisors is (1+1)(2+1)=(5+1)=6 .... for this , $p_i$ are prime numbers.

There is only one number of the first form and it is $2^5$

There are 15 numbers of the second form , as the prime which is squared can't be greater than 7 . Thus if $p_2=2$ then there are possibilities for $p_1$ as {3,5,7,11,13,17,19,23} hence 8 different numbers .

If $p_2=3$ then there are possibilities for $p_1$ as {2,5,7,11} hence 4 numbers.

If $p_2=5$ then $p_1$ can be {2,3} hence 2 numbers and for $p_2=7$ , $p_1=2$ .

Hence there are $1+8+4+2+1 = \boxed{16}$ such numbers

class dfghjkl { public static void main(String args[]) { int a,b,c,d=0; for(a=1;a<=100;a++) { c=0; for(b=1;b<=a;b++) { if(a%b==0) c++; } if(c==6) d++; } System.out.println(d); } }