1. ## 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?

2. 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?

3. 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 $\displaystyle 10^{n}$

4. 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?

5. 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?

6. 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.

7. 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.

8. Thanks so much for your explanation. I think I understand it now.