What is a number that has 5 divisors including 1 and itself?

Answer 1

If it is assumed that, by saying "divisor" only positive divisors are involved; it may be, "any Prime number" to the power of four. To find the number of the positive divisors of a definite number, we can follow these steps;

1- First, the number should be written in the form: A= ax.by.cz... where; a, b & c are prime numbers. For instance, 12 can be written as: 12=22.31

2- Number of positive divisors are; (x+1).(y+1).(z+1), an equation, which can be found by solving the related probability problem about the possible different combinations of multiplications to form the related number. So, 12 has; (2+1).(1+1)=6 divisors including 1 and itself, which are; 12, 6, 4, 3, 2, 1.

Returning to real problem, since the desired number of positive divisors are 5, equation 2 can only be written as; (4+1).(0+1)=5 So, x=4 & y=0, z=0...

So, the number is; A= a4 i.e; 16, 81, 625 etc...