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HSC 2015 MX2 Marathon (archive) (6 Viewers)

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Drsoccerball

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Re: HSC 2015 4U Marathon

New question !


 
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Sy123

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Re: HSC 2015 4U Marathon

can sum 1 explain
First of all,

Now, if we want to find a number that divides into , we need to be able to construct this number by multiplying 3's and 5's. What I mean is,

So, the problem then, can be translated into:



(at least for just a visualization). Note that the first line of this post is important because it means that any selection must be unique.

To do this problem now, see that there are (n+1) possible number of ways to pick n blue balls (either pick 0 balls, 1, 2, ..., n) and (m+1) possible number of ways to pick m red balls (0, 1, 2, ... , m). So the number of ways in total is (n+1)(m+1)

-----

Note that the translation into a problem about balls is purely to restate the problem in familiar terms, it is in no way essential to the solution.
 

Drsoccerball

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Re: HSC 2015 4U Marathon

First of all,

Now, if we want to find a number that divides into , we need to be able to construct this number by multiplying 3's and 5's. What I mean is,

So, the problem then, can be translated into:



(at least for just a visualization). Note that the first line of this post is important because it means that any selection must be unique.

To do this problem now, see that there are (n+1) possible number of ways to pick n blue balls (either pick 0 balls, 1, 2, ..., n) and (m+1) possible number of ways to pick m red balls (0, 1, 2, ... , m). So the number of ways in total is (n+1)(m+1)

-----

Note that the translation into a problem about balls is purely to restate the problem in familiar terms, it is in no way essential to the solution.
If the question stated that all integers excluding itself would the answer be nm?
 

Drsoccerball

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Re: HSC 2015 4U Marathon

im assuming this works with n amount of factors aswell?
 
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