1. ## Product of factors

In how many ways can 2310 be expressed as a product of 3 factors ? ( This will not involve Combinatorics analysis !)

2. ## Re: Product of factors

Hello, SheekhKebab!

In how many ways can 2310 be expressed as a product of 3 factors?
(This will not involve Combinatorics analysis!) . It won't?

I suppose a brute-force LIST is acceptable . . .

. . $\begin{array}{c}2\cdot3\cdot385 \\ 2\cdot5\cdot231 \\ 2\cdot7\cdot165 \\ 2\cdot11\cdot105 \\ 3\cdot5\cdot154 \\ 3\cdot7\cdot110 \\ 3\cdot11\cdot70 \\ 5\cdot7\cdot66 \\ 5\cdot11\cdot42 \\ 7\cdot11\cdot30 \end{array}$

3. ## Re: Product of factors

Originally Posted by Soroban

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I suppose a brute-force LIST is acceptable . . .
Hi Sir,

long time, no see ! There must be some algebraic method to solve this without Combinatorics, which is also simple and easy. And by the way the correct answer is 41.

4. ## Re: Product of factors

Originally Posted by SheekhKebab
In how many ways can 2310 be expressed as a product of 3 factors ? ( This will not involve Combinatorics analysis !)
Restating the problem

Find all {x,y,z} such that x,y,z integers, xyz = 2310, and $1\leq x \leq y \leq z$

The number of solutions is 41

5. ## Re: Product of factors

Originally Posted by SheekhKebab
Hi Sir,

long time, no see ! There must be some algebraic method to solve this without Combinatorics, which is also simple and easy. And by the way the correct answer is 41.
It looks comprehensive enough to me. Are you sure that you are't counting solutions like 11 x 5 x 42? If so then you do need the combinitorics after all.

-Dan

6. ## Re: Product of factors

You've forgotten one of factors may be 1

7. ## Re: Product of factors

Originally Posted by Cartesius24
You've forgotten one of factors may be 1
also 11,14,15 is not in the list

8. ## Re: Product of factors

Originally Posted by Idea
Restating the problemFind all {x,y,z} such that x,y,z integers, xyz = 2310, and $1\leq x \leq y \leq z$The number of solutions is 41
Hi Idea,

9. ## Re: Product of factors

Originally Posted by SheekhKebab
Hi Idea,

same as Soroban brute force list except I included 1 as a possible factor, and the first two factors don't have to be prime numbers

such as 6 11 35 and 7 10 33 are included in the list

There are 16 triplets of the form {1,y,z} since there are 32 divisors of 2310

I don't know how but I sense that there is a way to count the remaining 25. Is that what you are after?

10. ## Re: Product of factors

Originally Posted by Idea
I don't know how but I sense that there is a way to count the remaining 25. Is that what you are after?
Yep! I think the method without combinatorics will be similar to the method we use to express a number as a product of 2 factors. Also, I would like to know what will be the method under combinatorics for this question.

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### how many ways can 49 be express as product of two factors

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