SOLUTION: Find the number of positive integers that are divisors of at least one of 6^{6}, 10^{10}, 15^{15}, and 30^{30}.

Algebra.Com
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): You can put this solution on YOUR website!

Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!

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
RELATED QUESTIONS

Find the number of positive integers that are divisors of at least one of 10^10, 15^7,... (answered by Edwin McCravy)
A positive integer n has exactly 4 positive divisors that are perfect fifth powers,... (answered by Edwin McCravy)
If p and q are positive integers such that 7/10 < p/q < 11/15. then the smallest possible (answered by josgarithmetic,jim_thompson5910,solver91311,MathTherapy)
Find how many numbers in the range 1 - 300 are divisible by at least one of the integers... (answered by ikleyn)
Determine the number of ordered pairs of positive integers (a,b) such that a and b are... (answered by greenestamps)
Consider the experiment where a pair of fair dice is thrown. Let X denote the random (answered by stanbon)
Find the number of positive integers less than 1000 that contains at least one odd digit... (answered by AnlytcPhil)
what is product of the divisors of 1000 (including 1 and 1000)? a) 10^24 b) 10^16 (answered by Edwin McCravy)
The smallest positive integer with all of 2,3,4,5,8 and 10 as divisors is closest... (answered by richwmiller)