# positive integral divisors

• Sep 9th 2009, 04:29 PM
absvalue
positive integral divisors
The question is how many positive divisors do each of the following numbers have:

1) 6
2) 48
3) 100
4) $\displaystyle 2^{n}$
5) $\displaystyle 10^{n}$
6) 0

I understand how to do #1, #2, and #3. For example:

$\displaystyle 6 = 2^1 * 3^1 = (1 + 1) * (1 + 1) = 4$ positive divisors. (which happen to be 1, 2, 3, and 6)

I used the same method (converting each number to a factor of primes in exponential notation, adding 1 to each exponent, and multiplying the exponents) to solve #2 and #3.

I'm not sure where to start on 4, 5, or 6. Could anyone shed some light on the correct method for doing these problems?
• Sep 9th 2009, 04:47 PM
Plato
Quote:

Originally Posted by absvalue
The question is how many positive divisors do each of the following numbers have:
4) $\displaystyle 2^{n}$
5) $\displaystyle 10^{n}$
6) 0
I'm not sure where to start on 4, 5, or 6. Could anyone shed some light on the correct method for doing these problems?

For #4, there are $\displaystyle (n+1)$ factors. Why?

For #5, there are $\displaystyle (n+1)(n+1)$ factors. Why?

For #6, how many positive integers divide zero?
• Sep 9th 2009, 04:49 PM
pickslides
Quote:

Originally Posted by absvalue
The question is how many positive divisors do each of the following numbers have:

4) $\displaystyle 2^{n}$

I think the answer is n+1

You should be able to count them, try n = 1,2,3,4, and 5.

You'll see the pattern.

Try the same for $\displaystyle 10^{n}$
• Sep 9th 2009, 05:04 PM
absvalue
Quote:

Originally Posted by Plato
For #4, there are $\displaystyle (n+1)$ factors. Why?

For #5, there are $\displaystyle (n+1)(n+1)$ factors. Why?

For #6, how many positive integers divide zero?

Thanks, I think I see the pattern. That's beautiful.

For #4 there are n + 1 factors because:

$\displaystyle 2^{n} = (n + 1)$
$\displaystyle 2^{3} = (3 + 1) = 4$ (there are 4 positive divisors of 8)
$\displaystyle 2^{4} = (4 + 1) = 4$ (there are 5 positive divisors of 16)

For #5 there are $\displaystyle (n + 1)^2$ factors because:

$\displaystyle 10^{n} = (n + 1)^2$
$\displaystyle 10^2 = (2 + 1)^2 = 9$ (there are 9 positive divisors of 100)
$\displaystyle 10^3 = (3 + 1)^2 = 16$ (there are 16 positive divisors of 1000)

I understand the pattern above. But what if the question was a number other than $\displaystyle 2^{n}$ or $\displaystyle 10^{n}$? Is there a way to find the number of positive divisors for, say, $\displaystyle 8^{n}$?

For #6: Zero divided by any number (except zero, because that is undefined) should be zero. So the answer would be the set of all numbers not equal to zero?
• Sep 9th 2009, 05:12 PM
Plato
Here is what you need to know.
The number $\displaystyle 2^43^25^67$ has $\displaystyle (5)(3)(7)(2)$ divisors. Do you see why?

The number $\displaystyle 2^a3^b5^c7^d$ has $\displaystyle (a+1)(b+1)(c+1)(d+1)$ divisors. Do you see why?
• Sep 9th 2009, 05:44 PM
absvalue
Quote:

Originally Posted by Plato
Here is what you need to know.
The number $\displaystyle 2^43^25^67$ has $\displaystyle (5)(3)(7)(2)$ divisors. Do you see why?

The number $\displaystyle 2^a3^b5^c7^d$ has $\displaystyle (a+1)(b+1)(c+1)(d+1)$ divisors. Do you see why?

Yes, that makes sense. There are (5)(3)(7)(2) positive divisors because $\displaystyle 2^4$ has (4 + 1) divisors, $\displaystyle 3^2$ has (2 + 1) divisors, $\displaystyle 5^6$ has (6 + 1) divisors, and $\displaystyle 7^1$ has (1 + 1) divisors.

So, does this only work with prime numbers? Is it not possible to do this with $\displaystyle 16^{n}$ or $\displaystyle 8^{n}$ or $\displaystyle 20^{n}$?

Because $\displaystyle 8^{n}$ does not have $\displaystyle n + 1$ factors... For example:

$\displaystyle 8^2 = 64 = 2^6 = 6 + 1 = 7$ positive divisors.
• Sep 10th 2009, 06:47 AM
Plato
Quote:

Originally Posted by absvalue
So, does this only work with prime numbers? Is it not possible to do this with $\displaystyle 16^{n}$ or $\displaystyle 8^{n}$ or $\displaystyle 20^{n}$?

But every positive integer has a prime factorization.
Thus $\displaystyle 20^{n}=2^{2n}\cdot 5^n$. So the number has $\displaystyle (2n+1)(n+1)$ factors.
• Sep 10th 2009, 03:54 PM
absvalue
Thanks so much for your explanation. I think I understand it now. :)