Find all the positive divisors.

  • Thread starter cragar
  • Start date
  • #1
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Homework Statement


Find all the positive divisors of [itex] 10^n [/itex]
where n is a positive integer.

The Attempt at a Solution


can I factor this like [itex] 2^n5^n [/itex]
I notice that 10 has 4 divisors and 100 has 8
so it might seem that it has 4n divisors but im not should how to give a nice proof of it.
 

Answers and Replies

  • #2
22,089
3,297
Hi cragar! :smile:

Homework Statement


Find all the positive divisors of [itex] 10^n [/itex]
where n is a positive integer.

The Attempt at a Solution


can I factor this like [itex] 2^n5^n [/itex]
I notice that 10 has 4 divisors and 100 has 8
so it might seem that it has 4n divisors but im not should how to give a nice proof of it.

100 has more than 8 divisors! In fact, these are the divisors of 100:

[tex]1,2,4,5,10,20,25,50,100[/tex]

So 100 has 9 divisors! (you might see another pattern popping up now).

To prove the general conjecture, you need to find a connection between the divisors and the prime factorization. That is, if you know the prime factorization, how can you calculate all the divisors?

If you're unsure where to begin, begin by calculating the prime factorization of the 9 divisors of 100, and see if you notice something cool...
 
  • #3
360
0
Looks like a good start, cragar! Except 100 has 9 factors:
1, 2, 4, 5, 10, 20, 25, 50, 100.

Anyway, now it's essentially a combinatorics problem. Using your prime factorization of [itex] 10^n [/itex], any divisor will be of the form [itex] 2^k 5^k [/itex] with k between what and what? From there it's just a counting problem.

EDIT: Whoops, beaten to the punch!
 
  • #4
2,544
3
thanks for the help guys, I see that any divisor can be written as
[itex] 2^k5^d [/itex]
where k and d go from 0 to n , so i just have all the different possible combinations of k and d between 0 to n .
 
  • #5
22,089
3,297
And can you find how many different such combinations there are?
 
  • #6
2,967
5
You can have the following divisors, not sorted in an increasing order:

[tex]
\begin{array}{l}
1, 5, 5^{2}, \ldots, 5^{n} \\
2, 2*5, 2*5^{2}, \ldots, 2*5^{n} \\
\ldots \\
2^{n}, 2^{n}*5,2^{n}*5^{2}, \ldots, 2^{n}*5^{n}
\end{array}
[/tex]
 
  • #7
gb7nash
Homework Helper
805
1
Perhaps an easier way of looking at it is this:

Step 1: Pick number of 2's

n+1 ways (do you see why?)

Step 2: Pick number of 5's

n+1 ways (do you see why?)


Since these operations are independent...
 
  • #8
2,544
3
Perhaps an easier way of looking at it is this:

Step 1: Pick number of 2's

n+1 ways (do you see why?)

Step 2: Pick number of 5's

n+1 ways (do you see why?)
Since these operations are independent...
It seems like you would have to construct all the divisors from the primes from 0 to n,
im not sure exactly why you would pick them n+1 ways?
Is the reason i can pick the number of 2's (n+1) ways because I can pick it n ways and then I also want to include zero so I add one.
And then I will pick 5 the same way and then I will multiply my choices together
(n+1)(n+1)
 
Last edited:
  • #9
gb7nash
Homework Helper
805
1
Is the reason i can pick the number of 2's (n+1) ways because I can pick it n ways and then I also want to include zero so I add one.
And then I will pick 5 the same way and then I will multiply my choices together
(n+1)(n+1)

Correct
 
  • #10
2,544
3
sweet thanks
 

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