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|
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?
To return to the previous page use your browser's back button.