1. ## number of divisors

let $\displaystyle p_{i}$ be a prime number and $\displaystyle a_{i}$ be a positive integer for all $\displaystyle i$

a) list all distinct divisors of 32

1,2,4,8,16,32

$\displaystyle 2^5$ 5+1 = 6 divisors

b) how many distinct divisors does $\displaystyle p^a$ have?

a+1 divisors

c) how many distinct divisors does $\displaystyle p_{1}^{a_{1}}p_{2}^{a_{2}}\cdot\cdot\cdot p_{m}^{a_{m}}$ have?

$\displaystyle (a_{1}+1)(a_{2}+1)\cdot\cdot\cdot(a_{m}+1)$ divisors

d) how many distinct divisors does 60 have?

using prime factorization $\displaystyle 60 = 2^2\cdot3\cdot5$

so it has $\displaystyle (2+1)(1+1)(1+1) = 12$

here is the one I don't know how to solve

e) What is the smallest positive integer with exactly 100 distinct divisors?

2. ## Re: number of divisors

isn't it going to be the product of the first 100 primes?

3. ## Re: number of divisors

Originally Posted by romsek
isn't it going to be the product of the first 100 primes?

that seems to makes sense.

I was thinking I want $\displaystyle 100 = 2\cdot 50$ divisors

$\displaystyle 2 \cdot 50=2\cdot2\cdot5$

so the number would be something like $\displaystyle p^1q^1r^4$ but that doesn't really tell me what the number is.

I'm not really sure how to show it would be the product of the first 100 primes

4. ## Re: number of divisors

Originally Posted by Jonroberts74
that seems to makes sense.

I was thinking I want $\displaystyle 100 = 2\cdot 50$ divisors

$\displaystyle 2 \cdot 50=2\cdot2\cdot5$

so the number would be something like $\displaystyle p^1q^1r^4$ but that doesn't really tell me what the number is.

I'm not really sure how to show it would be the product of the first 100 primes
ok I read a similar problem. It's not the product of the 1st 100 primes.

you have

$n = \displaystyle{\prod_{k=1}^N}p_k^{a_k}$

You want to find $N$ and the $a_k$ such that

$\displaystyle{\prod_{k=1}^N}a_k=100$

and

$n$ is minimized.

I'll have to think about how to solve this one.

5. ## Re: number of divisors

Originally Posted by romsek
ok I read a similar problem. It's not the product of the 1st 100 primes.

you have

$n = \displaystyle{\prod_{k=1}^N}p_k^{a_k}$

You want to find $N$ and the $a_k$ such that

$\displaystyle{\prod_{k=1}^N}a_k=100$

and

$n$ is minimized.

I'll have to think about how to solve this one.
yeah I did some experimenting. I did the product of the first 7 primes and got 128 distinct divisors

6. ## Re: number of divisors

I was wrong

I said 2*2*5 = 50,

I meant 2*5*5= 50

so 2*2*5*5 equals 100

so the number should be

$\displaystyle p^1q^1r^4s^4$

so the first 4 primes

$\displaystyle 2^4 \times 3^4 \times 5^1 \times 7^1 = 45360$