The number of positive integral solutions is to be found

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The discussion centers on finding the number of positive integral solutions for the equation xyz=3000. Participants suggest factoring 3000 into its prime components and using combinatorial methods to approach the problem. One method mentioned involves applying the AM-GM inequality and the multinomial theorem. The conversation also touches on simplifying the problem by comparing it to distributing colored balls into buckets. Ultimately, the focus remains on determining the correct combinatorial approach to solve for x, y, and z.
ack
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I am preparing for an entrance and I came across this sum.The equation given is xyz=3000. we need to find how many positive integral solutions are there for x,y and z.
Please help.
 
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Let's factor 3000 into primes.
We are left with 23*3*53.

Now we are left with a very simple combinatorics question. Can you figure it out? Think about what items you are selecting and what you are placing them into.
 
I already tried that.x can be 2^0,2^1,2^2or2^3 that is 4 ways ,2ways for 3and 4ways for 5.But what about y?
I thought of another method ,
using AM>GM,

x+y+z>=43 and max can be 3002 (when one of them is 3000 nd the other two 1 each.)
Then use multinomial theorem.Can this be done?
 
ack said:
I already tried that.x can be 2^0,2^1,2^2or2^3 that is 4 ways ,2ways for 3and 4ways for 5.But what about y?
I thought of another method ,
using AM>GM,

x+y+z>=43 and max can be 3002 (when one of them is 3000 nd the other two 1 each.)
Then use multinomial theorem.Can this be done?
You are vastly overthinking this.
Consider the following problem: we have 3 red balls, 3 blue balls, and a green ball. How many different ways can we allocate them between 3 buckets?
 
Oh!...then the answer would be
9P7/(3!)^2.
thanks a lot!
 
Oh!...then the answer would be
9P7/(3!)^2.
thanks a lot!
 
Supposing we were to find all integral solutions?
 

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