What number has exactly 13 positive factors? Such as: 1x12=12, 2x6=12, etc.

When we say 'divisor' of a number we mean a second number that divides the first, for ex. 12 has divisors 1,2,3,4,6,12. Each of these (and only these)is a 'factor' of 12. This let's us write 12 as 1x2,2x6,3x4 - three different ways as a product of two factors. What you want is a number with exactly 13 divisors it would have to be a number like 2 raised to the 12th power. (4096). Let's see why:

We can write integers in terms of their prime factors, for example 12 = 2^2x3^1 (I'm using the ^ for the exponent here). Suppose we were to look for a divisor d of 12, then d must be made up of twos and threes as follows:

Number of twos	Number of threes	d
2	1	12
1	1	6
0	1	3
2	0	4
1	0	2
0	0	1

There are (2+1)x(1+1) = 6 cases -- three choices for the number of 2's times 2 choices for the number of 3's.

If a number N had a representation in terms of primes p & q, N = p^rxq^t, then each divisor d is made up of at most r p's and at most t q's giving (r+1)x(t+1) divisors in all. (In our example of N= 12 we have p=2, q=3, r=2, and t=1).

How many divisors would N = 2^3x3^5x5^2 have?

In your queation the number of divisors you wanted was an odd number (13). Can you guess what kind of numbers have an odd number of divisors?

Penny Nom