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# How many factors in total.

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For some positive integer $$n$$, the number $$110n^3$$ has $$110$$ positive integer divisors, including 1 and the number $$110n^3$$. How many positive integer divisors does the number $$81n^4$$ have?

Jun 19, 2021
edited by Melody  Jun 21, 2021

#8
+114044
+1

Ok I've got this.

https://www.quora.com/How-can-I-find-the-total-number-of-prime-factors-in-the-expression-4-11-x-7-5-x-11

From this question and answer I have ascertained the following.

Written in my words:

If a number is written as a product of its prime factors and those factors are presented in their power groupings then

if you add one to each of the powers and multiply them all together then you will get the number of factors in total.

for example:

$$2^3*7^1*19^2$$

will have 4*2*3 = 24 factors in total.

So now I will look at the question asked:

$$110=2*5*11$$      It is not relevant to the question but this will have 2*2*2 = 8 factors.

let a,b and c  =   2, 5 and 11   in any order.

We are told that    abc*n^3  will have 110 factors

And 110 = 2*5*11  (and they are all prime)

so  abc*n^3 can only have 3 prima factors.

So  n^3 must only have prime factors of a, b and c

The powers of those prime factors must be   2-1, 5-1, and 11-1  so that is   1, 4 and 10

so

$$abcn^3=ab^4c^{10}\\ abcn^3=abc*b^3c^{9}\\ abcn^3=abc*(bc^{3})^3\\ n=(bc^3)$$

It is important to note that b and c are different prime numbers and that and they are not equal to 3

so

$$81n^4=3^4(bc^3)^4\\ 81n^4=3^4*b^4*c^{12}\\$$

This will have  5*5*13 = 325 factors.

NOW I AM HAPPY !!

Jun 21, 2021

#1
+26122
+2

For some positive integer $$n$$, the number $$110n^3$$ has $$110$$ positive integer divisors,
including $$1$$ and the number $$110n^3$$.
How many positive integer divisors does the number $$81n^4$$ have?

$$\sigma_0(n)$$ is the divisor function

$$n=40: \begin{array}{|rcll|} \hline \sigma_0(110*40^3) &=& 110 \\ \sigma_0(81*40^4) &=& 325 \\ \hline \end{array}$$

Jun 19, 2021
#2
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Thanks Heureka,

Where did you get 40 from?

Melody  Jun 19, 2021
#3
+26122
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by trial and error with function $$\sigma_0(n)$$ in wolfram alpha

heureka  Jun 19, 2021
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ok thanks Heureka,

I think there should be a better way of doing it where n does not even have to be determined.

Or where it can be determined without trial and error....

Melody  Jun 19, 2021
#5
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There is 1 plus 3 prime factors of 110

1,2,5,11

so how many factors are there altogether     3C0+3C1+3C2+3C3 = 1+3+3+1 = 8

So how many factors are there for a number like  220?  where 2^2 is a factor ...

There must be a way to work this out easily.    I will think about it.

What I am getting at is that if we know that 110 has 8 divisors and we know 110n^3 has 110 divisors then we should be able to work out how many divisors n^3 has and how many of them are common with the 110 divisors.

Jun 19, 2021
#6
+2116
+1

I will give this my best shot.

110 = 2*5*11

I'm going to assume that n only has factors of 5 and 11.

n^3 = 5^a * 11^b

x = number of factors of 2 in 110n^3

y = number of factors of 5 in 110n^3

z = number of factors of 11 in 110n^3

(x+1)(y+1)(z+1) = number of factors in 110n^3

(2)(2+a)(2+b) = 110

a = 3, b = 9 or a = 9, b = 3

It doesn't really matter which one you choose.

n^3 = 5^9 * 11^3

n = 5^3 * 11

n^4 = 5^12 * 11^4

81*n^4 =  3^4 * 5^12 * 11^4

5*13*5 = 325

=^._.^=

Jun 19, 2021
#7
+114044
+1

Hi catmg,

It is your original assumption that I am having trouble with.

"I'm going to assume that n only has factors of 5 and 11. "

If there was no answer before yours, why would you assume this.

You do not need to answer.  (unless you really want to) I am just airing my thoughts.

Jun 20, 2021
#8
+114044
+1

Ok I've got this.

https://www.quora.com/How-can-I-find-the-total-number-of-prime-factors-in-the-expression-4-11-x-7-5-x-11

From this question and answer I have ascertained the following.

Written in my words:

If a number is written as a product of its prime factors and those factors are presented in their power groupings then

if you add one to each of the powers and multiply them all together then you will get the number of factors in total.

for example:

$$2^3*7^1*19^2$$

will have 4*2*3 = 24 factors in total.

So now I will look at the question asked:

$$110=2*5*11$$      It is not relevant to the question but this will have 2*2*2 = 8 factors.

let a,b and c  =   2, 5 and 11   in any order.

We are told that    abc*n^3  will have 110 factors

And 110 = 2*5*11  (and they are all prime)

so  abc*n^3 can only have 3 prima factors.

So  n^3 must only have prime factors of a, b and c

The powers of those prime factors must be   2-1, 5-1, and 11-1  so that is   1, 4 and 10

so

$$abcn^3=ab^4c^{10}\\ abcn^3=abc*b^3c^{9}\\ abcn^3=abc*(bc^{3})^3\\ n=(bc^3)$$

It is important to note that b and c are different prime numbers and that and they are not equal to 3

so

$$81n^4=3^4(bc^3)^4\\ 81n^4=3^4*b^4*c^{12}\\$$

This will have  5*5*13 = 325 factors.

NOW I AM HAPPY !!

Melody Jun 21, 2021