Question 1210186: Find the number of positive integers that are divisors of at least one of 6^{6}, 10^{10}, 15^{15}, and 30^{30}.
Found 6 solutions by CPhill, ikleyn, greenestamps, mccravyedwin, Edwin McCravy, AnlytcPhil:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem step by step.
**1. Prime Factorization**
* $6^6 = (2 \cdot 3)^6 = 2^6 \cdot 3^6$
* $10^{10} = (2 \cdot 5)^{10} = 2^{10} \cdot 5^{10}$
* $15^{15} = (3 \cdot 5)^{15} = 3^{15} \cdot 5^{15}$
* $30^{30} = (2 \cdot 3 \cdot 5)^{30} = 2^{30} \cdot 3^{30} \cdot 5^{30}$
**2. Understanding Divisors**
A divisor of any of these numbers will be of the form $2^a \cdot 3^b \cdot 5^c$, where:
* $0 \le a \le 30$ (since $2^{30}$ is the highest power of 2)
* $0 \le b \le 30$ (since $3^{30}$ is the highest power of 3)
* $0 \le c \le 30$ (since $5^{30}$ is the highest power of 5)
**3. Total Possible Divisors**
If we consider all possible combinations of a, b, and c, we have:
* 31 choices for a (0 to 30)
* 31 choices for b (0 to 30)
* 31 choices for c (0 to 30)
Therefore, the total number of possible divisors is $31 \cdot 31 \cdot 31 = 31^3 = 29791$.
**4. Inclusion-Exclusion Principle**
We need to find the number of divisors that are divisors of *at least one* of the given numbers. We can use the inclusion-exclusion principle.
* Let A be the set of divisors of $6^6$.
* Let B be the set of divisors of $10^{10}$.
* Let C be the set of divisors of $15^{15}$.
* Let D be the set of divisors of $30^{30}$.
We want to find $|A \cup B \cup C \cup D|$.
Since $30^{30}$ contains all prime factors of the other 3, any divisor of $6^6$, $10^{10}$, or $15^{15}$ will also be a divisor of $30^{30}$.
Therefore, $|A \cup B \cup C \cup D| = |D|$.
* $|D| = (30+1)(30+1)(30+1) = 31^3 = 29791$
**5. Final Answer**
The number of positive integers that are divisors of at least one of the given numbers is $31^3 = 29791$.
**Therefore, the answer is 29791.**
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
Find the number of positive integers that are divisors of at least one of 6^{6}, 10^{10}, 15^{15}, and 30^{30}.
~~~~~~~~~~~~~~~~~~~~~~~~
Notice that all three numbers  ,  and  are divisors of the number  .
Therefore, you can simplify the problem's formulation: you ONLY need to find the number of divisors of .
The prime decomposition for 30 is 30 = 2*3*5.
So, =  .
For this number, its divisors are all
1, 2,  , . . . ,  (31 divisor),
1, 3,  , . . . ,  (31 divisor),
1, 5,  , . . . ,  (31 divisor)
and all of their products.
The number of all their distinct products is 31*31*31 =  = 29791.
It is the ANSWER to the problem's question.
Solved.
I wrote this my post after @CPhill to provide the most simple straightforward solution/explanation
and to remove all unnecessary complications from the post by @CPhill.
/////////////////////////////
Edwin, I think it was very good from your side to make this editing in your post
and to exclude/ (to remove) your unjustified insinuations against me from the original version.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Note that 6^6, 10^10, and 15^15 are all factors of 30^30. It is therefore only necessary to find the number of positive integers that are factors of 30^30.
The standard method for determining the number of factors of a given integer N is
(1) find the prime factorization of N;
(2) add 1 to each exponent in the prime factorization; and
(3) multiply the numbers from step (2)
The prime factorization of 30 is 2*3*5, so the prime factorization of 30^30 is (2^30)(3^30)(5^30).
The number of factors of 30^30 is (30+1)(30+1)(30+1) = 31^3 = 29791
ANSWER: 29791
Answer by mccravyedwin(407)  (Show Source):
Answer by Edwin McCravy(20055)  (Show Source):
Answer by AnlytcPhil(1806)  (Show Source):
You can put this solution on YOUR website!
I think everybody has it right now. There has been some editing going on.
 .
All factors have to be of the form 
where there are 31 choices for each ordered triple of exponents (m,n,p),
namely 0,1,2,...30.
Edwin
|
|
|
